Class 10 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Textbook of Mathematics Grade 10 (FBISE / NBF)

Mastery Guide: Quadratic Equations - Standard Forms, Solution Methods, Nature of Roots, Symmetric Functions & Simultaneous Modeling

📖 Chapter 2: Quadratic Equations 📅 Updated: Sep 20, 2026

FBISE Class 10 • Single National Curriculum

Mastery Guide: Quadratic Equations

Standard Canonical Forms, Solution Methodologies, Reducible Classes, Discriminant Theory, Symmetric Functions, Simultaneous Systems & Real-World Kinematic Modeling

📖 Unit Overview & Target Learning Outcomes

Chapter 2 establishes the core algebraic foundation of polynomial equations of degree 2, bridging basic factoring with advanced analysis of roots, transformations, and nonlinear modeling. Under the Federal Board (FBISE) and Single National Curriculum (SNC) standards, students master:

Standard & Pure Forms: Expressing any second-degree equation in canonical standard form $ax^2 + bx + c = 0$ ($a \neq 0$) and identifying pure forms where $b = 0$.
Three Core Solution Techniques: Solving quadratic equations with complete proficiency via Factoring (Splitting Middle Term), Completing the Square, and applying the universally derived Quadratic Formula.
Equations Reducible to Quadratic Form: Recognizing and solving 5 distinct structural classes (Biquadratic $ax^4+bx^2+c=0$, Reciprocal/Rational $p(x) + \frac{1}{p(x)}$, Symmetric Reciprocals $x^2 + \frac{1}{x^2}$, Exponential $a \cdot k^{2x} + b \cdot k^x + c=0$, and Quartic Linear Products $(x+a)(x+b)(x+c)(x+d)=k$).
The Discriminant $\Delta = b^2 - 4ac$: Calculating the discriminant and determining the exact 4-case nature of roots (Rational & Unequal, Irrational & Unequal, Real & Equal, or Imaginary Conjugates).
Relations between Roots and Coefficients (Vieta's Formulas): Evaluating fundamental symmetric functions ($S = \alpha+\beta = -b/a$, $P = \alpha\beta = c/a$, $\alpha^2+\beta^2$, $\alpha^3+\beta^3$, etc.) and constructing new quadratic equations with transformed roots via $x^2 - Sx + P = 0$.
Complex Cube Roots of Unity: Deriving $1, \omega, \omega^2$ and applying the core cyclic identities $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$ to evaluate high-order powers and expressions.
Simultaneous Quadratic Systems: Solving systems of two equations in two variables (one linear + one quadratic, two pure quadratic, or homogeneous quadratic equations).
Applied Real-World Modeling: Formulating and solving word problems in physics kinematics ($S = v_i t + \frac{1}{2}at^2$), geometry (borders, areas, hypotenuses), and business revenue optimization.

💡 Kid-Friendly Tips for Success & Memory Hooks

1. The "Halve, Square, Add" Rule

When completing the square on $x^2 + bx$, take half of the middle coefficient $(b/2)$, square it $(b/2)^2$, and add it to BOTH sides of the equation!

2. Vieta's Sign Flip Trap

Remember that the Sum of Roots FLIPS sign: $S = \alpha+\beta = \mathbf{-b/a}$, but the Product KEEPS its sign: $P = \alpha\beta = \mathbf{c/a}$.

3. Discriminant Quick Check

Think of $\Delta = b^2 - 4ac$ as a traffic light: $\Delta > 0$ = Two Roads (2 Real Roots); $\Delta = 0$ = Single Bridge (1 Equal Root); $\Delta < 0$ = Flying into Imaginary Skies (2 Complex Roots)!

4. Omega ($\omega$) Power Reducer

To simplify any gigantic power $\omega^n$, divide $n$ by $3$ and keep ONLY the remainder: $\omega^{3k+r} = (\omega^3)^k \cdot \omega^r = 1 \cdot \omega^r = \mathbf{\omega^r}$!

🌍 Real-World Connections & Practical Engineering

Ballistics & Rocket Trajectories: In military aerospace and sports physics, gravity exerts constant deceleration $g = 9.8\text{ m/s}^2$. The altitude of any thrown ball or artillery projectile follows $h(t) = -\frac{1}{2}gt^2 + v_i t + h_0$, an exact quadratic parabola whose maximum height occurs at the vertex $t = v_i/g$.
Architectural Arches & Suspension Bridges: Suspension bridge cables and load-bearing monumental arches (like the Gateway Arch) distribute compressive stress according to quadratic and catenary parabolic curves $y = ax^2 + c$.
Satellite Dishes & Car Headlights: The reflective property of parabolas ensures that all incoming parallel rays focus exactly at a single focal point $(h, k + \frac{1}{4a})$, enabling high-gain telecommunications and intense headlight beam projection.
Business Profit & Pricing Models: When price increases, demand decreases linearly ($P = P_0 - kx$). Total revenue $R = \text{Price} \times \text{Quantity}$ creates a quadratic curve where the vertex pinpointing maximum profit can be found without calculus.

🔑 Study Cues & Essential Conceptual Inquiries

Q1: Why must the leading coefficient $a \neq 0$ in a quadratic equation?
Insight: If $a = 0$, the second-degree term $0 \cdot x^2$ vanishes, collapsing the equation into a linear equation $bx + c = 0$ (degree 1) which has at most 1 solution instead of 2.

Q2: Why does an equation reducible to quadratic form of degree 4 (like $x^4 - 5x^2 + 4 = 0$) yield 4 roots instead of 2?
Insight: By the Fundamental Theorem of Algebra, a degree 4 polynomial has 4 complex roots. Setting $y = x^2$ creates a quadratic in $y$ yielding two values ($y_1, y_2$). Then, solving $x^2 = y_1$ gives $\pm \sqrt{y_1}$ (2 roots) and $x^2 = y_2$ gives $\pm \sqrt{y_2}$ (2 roots), producing 4 solutions in total.

Q3: How do the discriminant $\Delta$ and the graphical x-intercepts of $y = ax^2+bx+c$ relate?
Insight: The number of real x-intercepts is determined completely by $\Delta$. If $\Delta > 0$, the parabola crosses the x-axis twice; if $\Delta = 0$, the vertex rests exactly on the x-axis (1 touching point); if $\Delta < 0$, the parabola floats entirely above or below the x-axis (0 real x-intercepts).

🌟 Section-by-Section Theory, Geometric Visuals & Core Formulations

2.1 Canonical Quadratic Forms & The Three Solution Methods

A second-degree polynomial equation in a single variable $x$ is termed a quadratic equation. In its canonical standard format, all non-zero terms are arranged in descending order of powers on the left-hand side: $$ax^2 + bx + c = 0 \quad (a, b, c \in \mathbb{R}, \, a \neq 0)$$ If $b = 0$, the linear term disappears, resulting in a pure quadratic equation $ax^2 + c = 0$.

📋 Table 1: Classification of Quadratic Equation Forms & Canonical Structures
Form Type	Algebraic Definition	Coefficients Constraint	Illustrative Textbook Examples	Solving Strategy
Standard Quadratic Form	$ax^2 + bx + c = 0$	$a \neq 0, b \neq 0, c \neq 0$	$2x^2 + 7x - 4 = 0$, $x^2 - 5x + 6 = 0$	Factoring, Completing the Square, or Quadratic Formula
Pure Quadratic Form	$ax^2 + c = 0$	$a \neq 0, b = 0$	$4x^2 - 9 = 0$, $3x^2 - 27 = 0$	Direct Square Root: $x = \pm \sqrt{-c/a}$
Monic Quadratic Form	$x^2 + bx + c = 0$	$a = 1$	$x^2 + 6x + 8 = 0$	Direct Factoring $(x+p)(x+q)=0$ or Completing Square
Incomplete Linear Missing Constant	$ax^2 + bx = 0$	$a \neq 0, c = 0$	$3x^2 + 7x = 0$, $5x^2 - 20x = 0$	Factoring common $x$: $x(ax+b)=0 \implies x=0, -b/a$

There are three universal algebraic methods to solve quadratic equations:

📋 Table 2: Comparative Analysis of Quadratic Solution Methods
Method Name	Algorithmic Procedure	Best Suited When	Speed & Efficiency	Limitations / Pitfalls
Factoring (Splitting Middle Term)	Find $p, q$ such that $p+q=b, pq=ac$; group & apply Zero Product Rule	$ac$ has easily identifiable integer factors	⚡ Ultra-fast for factorable integers	Fails or becomes inefficient for irrational/complex roots
Completing the Square	Divide by $a$, isolate constant, add $(b/2a)^2$ to both sides, take square root	Deriving vertex form $a(x-h)^2+k$ and proving formulas	Moderate, multi-step algebraic procedure	Fractional arithmetic can introduce sign/arithmetic slips
Quadratic Formula	Direct evaluation: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$	Any quadratic equation, especially with irrational/complex roots	Reliable, standardized formulaic approach	Requires strict sign tracking for negative coefficients $(-b)$
Graphical Solution	Plot $y = ax^2+bx+c$, locate vertex & x-intercepts $(x, 0)$	Visualizing zeros, extrema, and real-world trajectories	Visual insight & real-world intuition	Approximate for non-integer roots unless plotted accurately

Figure 2.1: Geometric Visualization of Completing the Square for $x^2 + bx + (b/2)^2 = (x + b/2)^2$.

2.2 Graphing Quadratic Functions, Vertex Coordinates & Symmetries

The Cartesian graph of every quadratic function $f(x) = ax^2 + bx + c$ forms a smooth, symmetric U-shaped curve called a parabola. By completing the square, the function can be converted into its canonical Vertex Form: $$f(x) = a(x - h)^2 + k$$ where the vertex is located at $(h, k) = \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right) = \left(-\frac{b}{2a}, -\frac{\Delta}{4a}\right)$.

Figure 2.2: Geometric Anatomy of Quadratic Parabola $y = ax^2 + bx + c$ showing Vertex, Axis of Symmetry, and x-intercepts.

2.3 The 5 Classes of Equations Reducible to Quadratic Form

Many higher-degree polynomial, rational, exponential, and radical equations can be transformed into standard quadratic equations by an appropriate change of variable ($u$ or $y$).

📋 Table 3: Canonical Taxonomy of Equations Reducible to Quadratic Form
Equation Class	Standard Structural Template	Core Substitution	Transformed Quadratic Equation	Back-Substitution Step
Type 1: Biquadratic	$ax^4 + bx^2 + c = 0$	$y = x^2$	$ay^2 + by + c = 0$	$x = \pm \sqrt{y}$ (yielding 4 roots)
Type 2: Rational / Reciprocal	$a p(x) + \frac{b}{p(x)} = c$	$y = p(x)$	$ay^2 - cy + b = 0$	$p(x) = y_1, y_2$
Type 3: Symmetric Reciprocal	$ax^4 + bx^3 + cx^2 + bx + a = 0$	Divide by $x^2$; let $y = x + \frac{1}{x}$	$a(y^2 - 2) + by + c = 0$	$x + \frac{1}{x} = y \implies x^2 - yx + 1 = 0$
Type 4: Exponential	$a \cdot k^{2x} + b \cdot k^x + c = 0$	$y = k^x$	$ay^2 + by + c = 0$	$k^x = y \implies x = \log_k(y)$
Type 5: Linear Quartic Product	$(x+a)(x+b)(x+c)(x+d) = k$ where $a+b = c+d$	$y = x^2 + (a+b)x$	$(y + ab)(y + cd) = k$	$x^2 + (a+b)x = y_1, y_2$

2.4 The Discriminant ($\Delta$) & The 4-Case Nature of Roots Matrix

In the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, the radicand quantity $\Delta = b^2 - 4ac$ governs the arithmetic nature and geometric reality of the roots without explicitly evaluating them.

📋 Table 4: Discriminant ($\Delta$) Decision Matrix & Nature of Roots
Discriminant Value ($\Delta = b^2 - 4ac$)	Nature of Roots	Algebraic Solution Character	Graphical Behavior of $y=ax^2+bx+c$	Discriminant Sign
$\Delta > 0$ and a Perfect Square	Real, Rational, Unequal	$x = \frac{-b \pm k}{2a} \in \mathbb{Q}$ ($k \in \mathbb{Z}$)	Parabola cuts x-axis at 2 distinct rational points	Positive ($\Delta > 0$)
$\Delta > 0$ but NOT a Perfect Square	Real, Irrational, Unequal	$x = \frac{-b \pm \sqrt{\Delta}}{2a}$ (Surd conjugate pair)	Parabola cuts x-axis at 2 distinct irrational points	Positive ($\Delta > 0$)
$\Delta = 0$	Real, Rational, Equal (Repeated)	$x = -\frac{b}{2a}$ (Multiplicity 2)	Parabola is tangent to x-axis at vertex $(-b/2a, 0)$	Zero ($\Delta = 0$)
$\Delta < 0$	Imaginary (Complex Conjugates)	$x = \frac{-b \pm i\sqrt{\|\Delta\|}}{2a} \in \mathbb{C}$	Parabola lies entirely above or below x-axis (0 x-intercepts)	Negative ($\Delta < 0$)

Figure 2.3: Discriminant Decision Tree for Classifying the Nature of Roots of Any Quadratic Equation.

2.5 Symmetric Functions of Roots & Vieta's Relations

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, Vieta's formulas establish that: $$\text{Sum: } S = \alpha + \beta = -\frac{b}{a}, \qquad \text{Product: } P = \alpha\beta = \frac{c}{a}$$ A function $f(\alpha, \beta)$ is called symmetric if interchanging $\alpha$ and $\beta$ leaves the expression unchanged ($f(\alpha, \beta) = f(\beta, \alpha)$).

📋 Table 5: Fundamental Symmetric Functions of Quadratic Roots
Symmetric Function	Algebraic Formula in $\alpha, \beta$	Expressible in terms of $S = -b/a$ and $P = c/a$	Standard Value for $ax^2+bx+c=0$
Sum of Roots	$\alpha + \beta$	$S$	$-\frac{b}{a}$
Product of Roots	$\alpha\beta$	$P$	$\frac{c}{a}$
Sum of Squares	$\alpha^2 + \beta^2$	$(\alpha+\beta)^2 - 2\alpha\beta = S^2 - 2P$	$\frac{b^2 - 2ac}{a^2}$
Difference Squared	$(\alpha - \beta)^2$	$(\alpha+\beta)^2 - 4\alpha\beta = S^2 - 4P$	$\frac{b^2 - 4ac}{a^2} = \frac{\Delta}{a^2}$
Sum of Reciprocals	$\frac{1}{\alpha} + \frac{1}{\beta}$	$\frac{\alpha+\beta}{\alpha\beta} = \frac{S}{P}$	$-\frac{b}{c}$
Sum of Reciprocal Squares	$\frac{1}{\alpha^2} + \frac{1}{\beta^2}$	$\frac{\alpha^2+\beta^2}{(\alpha\beta)^2} = \frac{S^2 - 2P}{P^2}$	$\frac{b^2 - 2ac}{c^2}$
Sum of Cubes	$\alpha^3 + \beta^3$	$(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = S^3 - 3SP$	$-\frac{b(b^2 - 3ac)}{a^3}$
Ratio Sum	$\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$	$\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{S^2 - 2P}{P}$	$\frac{b^2 - 2ac}{ac}$

2.6 Complex Cube Roots of Unity ($1, \omega, \omega^2$)

Solving the cubic equation $x^3 = 1 \implies x^3 - 1 = (x-1)(x^2 + x + 1) = 0$ yields three distinct cube roots: one real root ($x = 1$) and two non-real complex conjugate roots: $$\omega = \frac{-1 + i\sqrt{3}}{2}, \qquad \omega^2 = \frac{-1 - i\sqrt{3}}{2}$$

📋 Table 6: Algebraic Properties of Complex Cube Roots of Unity
Property Description	Mathematical Statement	Algebraic Proof Outline	Key Deductions & Identities
Three Cube Roots of Unity	$1, \omega, \omega^2$	$x^3 - 1 = (x-1)(x^2+x+1) = 0 \implies x = 1, \frac{-1 \pm i\sqrt{3}}{2}$	$\omega = \frac{-1 + i\sqrt{3}}{2}, \omega^2 = \frac{-1 - i\sqrt{3}}{2}$
Each Complex Root is Square of Other	$(\omega)^2 = \omega^2$ and $(\omega^2)^2 = \omega$	$\left(\frac{-1+i\sqrt{3}}{2}\right)^2 = \frac{1 - 2i\sqrt{3} - 3}{4} = \frac{-1-i\sqrt{3}}{2} = \omega^2$	Symmetric cyclic pairing
Sum of All Cube Roots is Zero	$1 + \omega + \omega^2 = 0$	$1 + \left(\frac{-1+i\sqrt{3}}{2}\right) + \left(\frac{-1-i\sqrt{3}}{2}\right) = 1 - 1 = 0$	$1 + \omega = -\omega^2, 1 + \omega^2 = -\omega, \omega + \omega^2 = -1$
Product of All Cube Roots is One	$1 \cdot \omega \cdot \omega^2 = \omega^3 = 1$	$\omega \cdot \omega^2 = \left(\frac{-1+i\sqrt{3}}{2}\right)\left(\frac{-1-i\sqrt{3}}{2}\right) = \frac{1 - 3i^2}{4} = 1$	$\omega^{3k} = 1, \omega^{3k+1} = \omega, \omega^{3k+2} = \omega^2$

2.7 Simultaneous Quadratic Systems & Real-World Modeling

A system of equations involving at least one second-degree polynomial in two variables is a simultaneous quadratic system.

📋 Table 7: Classification of Simultaneous Quadratic Systems
System Classification	Algebraic Structure	Standard Solution Method	Maximum Number of Real Solutions
Case 1: One Linear & One Quadratic	$y = mx + c$ and $Ax^2 + By^2 + Cx + Dy + E = 0$	Substitute linear expression $y = mx+c$ into quadratic equation	Up to 2 solution pairs $(x, y)$
Case 2: Two Pure Quadratic Equations	$A_1 x^2 + B_1 y^2 = C_1$ and $A_2 x^2 + B_2 y^2 = C_2$	Eliminate one variable ($x^2$ or $y^2$) via linear combination	Up to 4 solution pairs $(\pm x, \pm y)$
Case 3: Homogeneous Quadratic System	$A_1 x^2 + B_1 xy + C_1 y^2 = D_1$ and $A_2 x^2 + B_2 xy + C_2 y^2 = D_2$	Eliminate constant terms to get $A x^2 + B xy + C y^2 = 0$; factor into 2 linear relations	Up to 4 solution pairs $(x, y)$

📋 Table 8: Real-World Engineering, Physics & Business Quadratic Formulations
Application Domain	Mathematical Model	Standard Quadratic Translation	Variables & Physical Interpretation
Projectile Motion (Kinematics)	$S(t) = v_i t + \frac{1}{2} a t^2$	$\frac{1}{2}gt^2 - v_i t + S = 0$	$t$: time, $v_i$: initial velocity, $g$: gravitational acceleration, $S$: height
Geometric Area & Borders	$\text{Area} = (L + 2x)(W + 2x)$	$4x^2 + 2(L+W)x + (LW - A_{\text{total}}) = 0$	$x$: uniform border width, $L, W$: interior dimensions
Revenue & Business Optimization	$R(x) = (P_0 - kx)(Q_0 + mx)$	$-km x^2 + (mP_0 - kQ_0)x + P_0 Q_0 - R = 0$	$x$: price adjustments, $R(x)$: total revenue parabolic curve
Work-Rate & Pipeline Filling	$\frac{1}{t_1} + \frac{1}{t_1 + d} = \frac{1}{T_{\text{total}}}$	$T_{\text{total}}(2t_1 + d) = t_1(t_1 + d) \implies t_1^2 + (d - 2T)t_1 - dT = 0$	$t_1$: pipe 1 time, $d$: time difference, $T_{\text{total}}$: joint time

🎯 Unit Synthesis Summary

Chapter 2 (Quadratic Equations) synthesizes the algebra of degree-2 polynomials into a unified framework connecting algebraic manipulation, analytical geometry, and physical kinematics. We established that every quadratic equation can be brought to standard form $ax^2 + bx + c = 0$ ($a \neq 0$) and solved via Factoring, Completing the Square, or the Quadratic Formula. The Discriminant $\Delta = b^2 - 4ac$ completely characterizes the nature of roots (rational, irrational, equal, or imaginary) and mirrors the number of x-intercepts on the parabolic curve. Vieta's formulas ($S = -b/a, P = c/a$) unlock powerful symmetric transformations and enable reverse-engineering quadratic equations ($x^2 - Sx + P = 0$). Finally, substitution techniques empower us to solve complex higher-degree, reciprocal, and exponential equations, while simultaneous systems and kinematics formulas translate real-world dimensions, trajectories, and financial optimizations into clean quadratic resolutions.

📝 Part 2: Solved Textbook Exercises (Comprehensive FBISE Solution Manual)

Exercise 2.1 • Standard Form, Factoring, Completing Square, Formula & Word Problems

Topic: Standard Form of Quadratic Equations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q1 (i)** Write the following quadratic equation in standard form $ax^2 + bx + c = 0$: $$(x+7)(x-3) = -7$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Expand the product): $(x+7)(x-3) = x^2 - 3x + 7x - 21 = x^2 + 4x - 21$.
• Step 2 (Shift constant to left): $x^2 + 4x - 21 + 7 = 0 \implies \mathbf{x^2 + 4x - 14 = 0}$.
• Final Answer: $\mathbf{x^2 + 4x - 14 = 0}$ (Standard quadratic form where $a=1, b=4, c=-14$).

Topic: Standard Form of Quadratic Equations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q1 (ii)** Write the following quadratic equation in standard form $ax^2 + bx + c = 0$: $$\frac{x^2+4}{3} - \frac{x}{7} = 1$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Clear denominators by multiplying by $\text{LCM}(3,7)=21$): $21 \left(\frac{x^2+4}{3}\right) - 21 \left(\frac{x}{7}\right) = 21(1)$.
• Step 2 (Simplify terms): $7(x^2+4) - 3(x) = 21 \implies 7x^2 + 28 - 3x = 21$.
• Step 3 (Rearrange into standard form): $7x^2 - 3x + 28 - 21 = 0 \implies \mathbf{7x^2 - 3x + 7 = 0}$.
• Final Answer: $\mathbf{7x^2 - 3x + 7 = 0}$

Topic: Standard Form of Quadratic Equations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q1 (iii)** Write the following quadratic equation in standard form $ax^2 + bx + c = 0$: $$\frac{x}{x+1} + \frac{x+1}{x} = 6$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Multiply the entire equation by $x(x+1)$): $x(x) + (x+1)^2 = 6x(x+1)$.
• Step 2 (Expand all terms): $x^2 + (x^2 + 2x + 1) = 6x^2 + 6x \implies 2x^2 + 2x + 1 = 6x^2 + 6x$.
• Step 3 (Collect all terms on one side): $6x^2 - 2x^2 + 6x - 2x - 1 = 0 \implies \mathbf{4x^2 + 4x - 1 = 0}$.
• Final Answer: $\mathbf{4x^2 + 4x - 1 = 0}$

Topic: Solving Quadratic Equations by Factoring FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q2 (i)** Solve the following quadratic equation by factorization: $$x^2 - x - 20 = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Find factors of product $-20$ with sum $-1$): Factors are $-5$ and $+4$.
• Step 2 (Split middle term and factor by grouping): $x^2 - 5x + 4x - 20 = 0 \implies x(x-5) + 4(x-5) = 0 \implies (x-5)(x+4) = 0$.
• Step 3 (Apply Zero Product Property): $x - 5 = 0 \implies x = 5$ or $x + 4 = 0 \implies x = -4$.
• Final Answer: $\text{Solution Set} = \mathbf{\{-4, 5\}}$

Topic: Solving Quadratic Equations by Factoring FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q2 (ii)** Solve the following quadratic equation by factorization: $$3y^2 = y(y-5)$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Expand and bring all terms to LHS): $3y^2 = y^2 - 5y \implies 3y^2 - y^2 + 5y = 0 \implies 2y^2 + 5y = 0$.
• Step 2 (Factor out common term $y$): $y(2y + 5) = 0$.
• Step 3 (Solve for $y$): $y = 0$ or $2y + 5 = 0 \implies y = -\frac{5}{2}$.
• Final Answer: $\text{Solution Set} = \mathbf{\left\{0, -\frac{5}{2}\right\}}$

Topic: Solving Quadratic Equations by Factoring FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q2 (iii)** Solve the following quadratic equation by factorization: $$4 - 32x = 17x^2$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Rearrange into standard form): $17x^2 + 32x - 4 = 0$.
• Step 2 (Find product $17 \times (-4) = -68$ with sum $+32$): Factors are $+34$ and $-2$.
• Step 3 (Factor by grouping): $17x^2 + 34x - 2x - 4 = 0 \implies 17x(x+2) - 2(x+2) = 0 \implies (17x - 2)(x + 2) = 0$.
• Step 4 (Solve for $x$): $x + 2 = 0 \implies x = -2$ or $17x - 2 = 0 \implies x = \frac{2}{17}$.
• Final Answer: $\text{Solution Set} = \mathbf{\left\{-2, \frac{2}{17}\right\}}$

Topic: Solving Quadratic Equations by Factoring FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q2 (iv)** Solve the following quadratic equation by factorization: $$x^2 - 11x = 152$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Standard form): $x^2 - 11x - 152 = 0$.
• Step 2 (Find product $-152$ with sum $-11$): Factors are $-19$ and $+8$ (since $-19 \times 8 = -152$).
• Step 3 (Factor): $(x - 19)(x + 8) = 0$.
• Step 4 (Solve): $x = 19$ or $x = -8$.
• Final Answer: $\text{Solution Set} = \mathbf{\{-8, 19\}}$

Topic: Solving Quadratic Equations by Factoring FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q2 (v)** Solve the following quadratic equation by factorization: $$\frac{x+1}{x} + \frac{x}{x+1} = \frac{25}{12}$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Clear denominators by multiplying by $12x(x+1)$): $12(x+1)^2 + 12x^2 = 25x(x+1)$.
• Step 2 (Expand and simplify): $12(x^2 + 2x + 1) + 12x^2 = 25x^2 + 25x \implies 24x^2 + 24x + 12 = 25x^2 + 25x$.
• Step 3 (Rearrange): $25x^2 - 24x^2 + 25x - 24x - 12 = 0 \implies x^2 + x - 12 = 0$.
• Step 4 (Factorize): $(x + 4)(x - 3) = 0 \implies x = 3$ or $x = -4$.
• Final Answer: $\text{Solution Set} = \mathbf{\{-4, 3\}}$

Topic: Completing the Square Method FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q3 (i)** Solve the following equation by completing the square method: $$7x^2 + 2x - 1 = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Divide by leading coefficient $7$): $x^2 + \frac{2}{7}x - \frac{1}{7} = 0 \implies x^2 + \frac{2}{7}x = \frac{1}{7}$.
• Step 2 (Add $\left(\frac{1}{2} \cdot \frac{2}{7}\right)^2 = \left(\frac{1}{7}\right)^2 = \frac{1}{49}$ to both sides): $x^2 + \frac{2}{7}x + \frac{1}{49} = \frac{1}{7} + \frac{1}{49}$.
• Step 3 (Complete the square on LHS): $\left(x + \frac{1}{7}\right)^2 = \frac{7+1}{49} = \frac{8}{49}$.
• Step 4 (Take square root on both sides): $x + \frac{1}{7} = \pm \frac{\sqrt{8}}{7} = \pm \frac{2\sqrt{2}}{7} \implies x = \frac{-1 \pm 2\sqrt{2}}{7}$.
• Final Answer: $\mathbf{x = \frac{-1 \pm 2\sqrt{2}}{7}}$

Topic: Completing the Square Method FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q3 (ii)** Solve the following equation by completing the square method: $$ax^2 + 4x - a = 0 \quad (a \neq 0)$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Divide by $a$): $x^2 + \frac{4}{a}x - 1 = 0 \implies x^2 + \frac{4}{a}x = 1$.
• Step 2 (Add $\left(\frac{2}{a}\right)^2 = \frac{4}{a^2}$ to both sides): $x^2 + \frac{4}{a}x + \frac{4}{a^2} = 1 + \frac{4}{a^2} = \frac{a^2+4}{a^2}$.
• Step 3 (Complete the square): $\left(x + \frac{2}{a}\right)^2 = \frac{a^2+4}{a^2}$.
• Step 4 (Take square root): $x + \frac{2}{a} = \pm \frac{\sqrt{a^2+4}}{a} \implies x = \frac{-2 \pm \sqrt{a^2+4}}{a}$.
• Final Answer: $\mathbf{x = \frac{-2 \pm \sqrt{a^2+4}}{a}}$

Topic: Completing the Square Method FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q3 (iii)** Solve the following equation by completing the square method: $$11x^2 - 34x + 3 = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Divide by $11$ and isolate constant): $x^2 - \frac{34}{11}x = -\frac{3}{11}$.
• Step 2 (Add $\left(\frac{17}{11}\right)^2 = \frac{289}{121}$ to both sides): $x^2 - \frac{34}{11}x + \frac{289}{121} = -\frac{3}{11} + \frac{289}{121} = \frac{-33 + 289}{121} = \frac{256}{121}$.
• Step 3 (Complete square): $\left(x - \frac{17}{11}\right)^2 = \left(\frac{16}{11}\right)^2$.
• Step 4 (Take square root): $x - \frac{17}{11} = \pm \frac{16}{11} \implies x = \frac{17+16}{11} = 3$ or $x = \frac{17-16}{11} = \frac{1}{11}$.
• Final Answer: $\text{Solution Set} = \mathbf{\left\{3, \frac{1}{11}\right\}}$

Topic: Completing the Square Method FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q3 (iv)** Solve the following equation by completing the square method: $$lx^2 + mx + n = 0 \quad (l \neq 0)$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Divide by $l$): $x^2 + \frac{m}{l}x = -\frac{n}{l}$.
• Step 2 (Add $\left(\frac{m}{2l}\right)^2 = \frac{m^2}{4l^2}$): $x^2 + \frac{m}{l}x + \frac{m^2}{4l^2} = \frac{m^2}{4l^2} - \frac{n}{l} = \frac{m^2 - 4ln}{4l^2}$.
• Step 3 (Complete square and take square root): $\left(x + \frac{m}{2l}\right)^2 = \frac{m^2-4ln}{4l^2} \implies x + \frac{m}{2l} = \pm \frac{\sqrt{m^2-4ln}}{2l}$.
• Step 4 (Solve for $x$): $x = \frac{-m \pm \sqrt{m^2-4ln}}{2l}$.
• Final Answer: $\mathbf{x = \frac{-m \pm \sqrt{m^2-4ln}}{2l}}$

Topic: Completing the Square Method FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q3 (v)** Solve the following equation by completing the square method: $$3x^2 + 7x = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Divide by $3$): $x^2 + \frac{7}{3}x = 0$.
• Step 2 (Add $\left(\frac{7}{6}\right)^2 = \frac{49}{36}$): $x^2 + \frac{7}{3}x + \frac{49}{36} = \frac{49}{36}$.
• Step 3 (Factor and take square root): $\left(x + \frac{7}{6}\right)^2 = \left(\frac{7}{6}\right)^2 \implies x + \frac{7}{6} = \pm \frac{7}{6}$.
• Step 4 (Calculate roots): $x = -\frac{7}{6} + \frac{7}{6} = 0$ or $x = -\frac{7}{6} - \frac{7}{6} = -\frac{14}{6} = -\frac{7}{3}$.
• Final Answer: $\text{Solution Set} = \mathbf{\left\{0, -\frac{7}{3}\right\}}$

Topic: Completing the Square Method FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q3 (vi)** Solve the following equation by completing the square method: $$(x+2)^2 - 3x = 10$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Expand and write in standard form): $x^2 + 4x + 4 - 3x - 10 = 0 \implies x^2 + x - 6 = 0 \implies x^2 + x = 6$.
• Step 2 (Add $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ to both sides): $x^2 + x + \frac{1}{4} = 6 + \frac{1}{4} = \frac{25}{4}$.
• Step 3 (Complete square and solve): $\left(x + \frac{1}{2}\right)^2 = \frac{25}{4} \implies x + \frac{1}{2} = \pm \frac{5}{2}$.
• Step 4 (Calculate roots): $x = -\frac{1}{2} + \frac{5}{2} = 2$ or $x = -\frac{1}{2} - \frac{5}{2} = -3$.
• Final Answer: $\text{Solution Set} = \mathbf{\{-3, 2\}}$

Topic: Quadratic Formula Application FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q4 (i)** Solve using the quadratic formula: $$2 - x^2 = 7x$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Standard form): Shift terms to RHS: $x^2 + 7x - 2 = 0$. Here $a=1, b=7, c=-2$.
• Step 2 (Apply Quadratic Formula): $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4(1)(-2)}}{2(1)}$.
• Step 3 (Simplify under radical): $\Delta = 49 + 8 = 57 \implies x = \frac{-7 \pm \sqrt{57}}{2}$.
• Final Answer: $\mathbf{x = \frac{-7 \pm \sqrt{57}}{2}}$

Topic: Quadratic Formula Application FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q4 (ii)** Solve using the quadratic formula: $$5x^2 + 8x + 1 = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Identify coefficients): $a=5, b=8, c=1$.
• Step 2 (Apply Quadratic Formula): $x = \frac{-8 \pm \sqrt{8^2 - 4(5)(1)}}{2(5)} = \frac{-8 \pm \sqrt{64 - 20}}{10} = \frac{-8 \pm \sqrt{44}}{10}$.
• Step 3 (Simplify radical): $\sqrt{44} = 2\sqrt{11} \implies x = \frac{-8 \pm 2\sqrt{11}}{10} = \frac{2(-4 \pm \sqrt{11})}{10} = \frac{-4 \pm \sqrt{11}}{5}$.
• Final Answer: $\mathbf{x = \frac{-4 \pm \sqrt{11}}{5}}$

Topic: Quadratic Formula Application FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q4 (iii)** Solve using the quadratic formula: $$\sqrt{3}x^2 + x = 4\sqrt{3}$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Standard form): $\sqrt{3}x^2 + x - 4\sqrt{3} = 0$. Here $a=\sqrt{3}, b=1, c=-4\sqrt{3}$.
• Step 2 (Calculate Discriminant): $\Delta = b^2 - 4ac = 1^2 - 4(\sqrt{3})(-4\sqrt{3}) = 1 + 4(12) = 1 + 48 = 49$.
• Step 3 (Apply Formula): $x = \frac{-1 \pm \sqrt{49}}{2\sqrt{3}} = \frac{-1 \pm 7}{2\sqrt{3}}$.
• Step 4 (Evaluate roots): $x_1 = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$; $x_2 = \frac{-8}{2\sqrt{3}} = \frac{-4}{\sqrt{3}} = -\frac{4\sqrt{3}}{3}$.
• Final Answer: $\text{Solution Set} = \mathbf{\left\{\sqrt{3}, -\frac{4}{\sqrt{3}}\right\}}$

Topic: Quadratic Formula Application FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q4 (iv)** Solve using the quadratic formula: $$4x^2 - 14 = 3x$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Standard form): $4x^2 - 3x - 14 = 0$. Here $a=4, b=-3, c=-14$.
• Step 2 (Calculate Discriminant): $\Delta = (-3)^2 - 4(4)(-14) = 9 + 224 = 233$.
• Step 3 (Apply Formula): $x = \frac{-(-3) \pm \sqrt{233}}{2(4)} = \frac{3 \pm \sqrt{233}}{8}$.
• Final Answer: $\mathbf{x = \frac{3 \pm \sqrt{233}}{8}}$

Topic: Quadratic Formula Application FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q4 (v)** Solve using the quadratic formula: $$6x^2 - 3 - 7x = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Standard form): $6x^2 - 7x - 3 = 0$. Here $a=6, b=-7, c=-3$.
• Step 2 (Discriminant): $\Delta = (-7)^2 - 4(6)(-3) = 49 + 72 = 121$.
• Step 3 (Solve): $x = \frac{7 \pm \sqrt{121}}{12} = \frac{7 \pm 11}{12}$.
• Step 4 (Roots): $x_1 = \frac{18}{12} = \frac{3}{2}$; $x_2 = \frac{-4}{12} = -\frac{1}{3}$.
• Final Answer: $\text{Solution Set} = \mathbf{\left\{\frac{3}{2}, -\frac{1}{3}\right\}}$

Topic: Quadratic Formula Application FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q4 (vi)** Solve using the quadratic formula: $$3x^2 + 8x + 2 = 0$$

Exhaustive Step-by-Step Resolution:

• Step 1 (Identify coefficients): $a=3, b=8, c=2$.
• Step 2 (Discriminant): $\Delta = 8^2 - 4(3)(2) = 64 - 24 = 40$.
• Step 3 (Solve): $x = \frac{-8 \pm \sqrt{40}}{6} = \frac{-8 \pm 2\sqrt{10}}{6} = \frac{-4 \pm \sqrt{10}}{3}$.
• Final Answer: $\mathbf{x = \frac{-4 \pm \sqrt{10}}{3}}$

Topic: Graphing and Factoring Quadratic Functions FBISE Rubric: 8 Marks (LONG)

**Exercise 2.1 Q5** Solve the equation $x^2 + 6x = -9$ both by factoring and by graphing the corresponding quadratic function $y = x^2 + 6x + 9$. Interpret the geometric significance of the root.

Exhaustive Step-by-Step Resolution:

• Part A (Algebraic Factoring):
1. Write in standard form: $x^2 + 6x + 9 = 0$.
2. Recognize perfect square trinomial: $(x+3)^2 = 0 \implies (x+3)(x+3) = 0$.
3. By Zero Product Rule: $x = -3$ (repeated root).
• Part B (Table of Values for Graphing $y = x^2 + 6x + 9$):
- For $x = -5$: $y = (-5)^2 + 6(-5) + 9 = 25 - 30 + 9 = 4$.
- For $x = -4$: $y = 16 - 24 + 9 = 1$.
- For $x = -3$: $y = 9 - 18 + 9 = 0$ → Vertex & x-intercept.
- For $x = -2$: $y = 4 - 12 + 9 = 1$.
- For $x = -1$: $y = 1 - 6 + 9 = 4$.
• Geometric Interpretation: The parabola opens upward with its vertex located exactly at $(-3, 0)$. Because the parabola touches (is tangent to) the x-axis at a single point, the equation has exactly one distinct real root (a repeated root of multiplicity 2).
• Final Answer: $\mathbf{x = -3}$

Topic: Vertex Form and Graphing Quadratic Functions FBISE Rubric: 8 Marks (LONG)

**Exercise 2.1 Q6** For the quadratic function $y = x^2 + 2x + 4$, find its vertex by completing the square, plot the curve, and explain why $x^2 + 2x + 4 = 0$ has no real solutions.

Exhaustive Step-by-Step Resolution:

• Step 1 (Convert to Vertex Form by Completing the Square):
$y = (x^2 + 2x) + 4 = (x^2 + 2x + 1) + 4 - 1 = (x + 1)^2 + 3$.
Comparing with vertex form $y = a(x-h)^2 + k$, the vertex is $(h, k) = \mathbf{(-1, 3)}$.
• Step 2 (Calculate Discriminant of $x^2 + 2x + 4 = 0$):
$\Delta = b^2 - 4ac = 2^2 - 4(1)(4) = 4 - 16 = -12 < 0$.
• Step 3 (Geometric & Algebraic Interpretation):
Since the leading coefficient $a=1 > 0$, the parabola opens upward, and its absolute minimum value is $y=3$ (at $x=-1$). The graph never crosses or touches the x-axis ($y=0$), confirming geometrically that there are no real roots (the roots are complex conjugates: $x = -1 \pm i\sqrt{3}$).
• Final Answer: $\mathbf{\text{Vertex } (-1, 3); \text{No real roots.}}$

Topic: Fundamental Definitions of Quadratic Theory FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q7 (i)** Define the mathematical term: **Solution**.

Exhaustive Step-by-Step Resolution:

• Definition: Solution: A value of the variable that satisfies a given equation, making the left-hand side equal to the right-hand side.
• Mathematical Relationship: The zeros of the function $f(x) = ax^2+bx+c$ correspond to the roots of the equation $ax^2+bx+c=0$ and the x-intercepts $(x, 0)$ on its Cartesian graph.

Topic: Fundamental Definitions of Quadratic Theory FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q7 (ii)** Define the mathematical term: **Root of an equation**.

Exhaustive Step-by-Step Resolution:

• Definition: Root of an equation: A specific numerical value of $x$ for which the equation $P(x) = 0$ becomes true.
• Mathematical Relationship: The zeros of the function $f(x) = ax^2+bx+c$ correspond to the roots of the equation $ax^2+bx+c=0$ and the x-intercepts $(x, 0)$ on its Cartesian graph.

Topic: Fundamental Definitions of Quadratic Theory FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q7 (iii)** Define the mathematical term: **Zero of a function**.

Exhaustive Step-by-Step Resolution:

• Definition: Zero of a function: An input value $x$ in the domain of a function $f$ such that the output $f(x) = 0$.
• Mathematical Relationship: The zeros of the function $f(x) = ax^2+bx+c$ correspond to the roots of the equation $ax^2+bx+c=0$ and the x-intercepts $(x, 0)$ on its Cartesian graph.

Topic: Fundamental Definitions of Quadratic Theory FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q7 (iv)** Define the mathematical term: **x-intercept**.

Exhaustive Step-by-Step Resolution:

• Definition: x-intercept: The coordinate point $(x, 0)$ where the graph of a function crosses or touches the horizontal x-axis.
• Mathematical Relationship: The zeros of the function $f(x) = ax^2+bx+c$ correspond to the roots of the equation $ax^2+bx+c=0$ and the x-intercepts $(x, 0)$ on its Cartesian graph.

Topic: Fundamental Theorem of Algebra FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.1 Q8** Can a quadratic equation have more than two solutions? Justify your answer mathematically.

Exhaustive Step-by-Step Resolution:

• Explanation: By the Fundamental Theorem of Algebra, any polynomial equation $P(x) = 0$ of degree $n$ with complex coefficients has exactly $n$ complex roots (counting multiplicity).
• For a quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$), the degree is $n = 2$. Therefore, it can have at most two distinct roots (or one repeated root of multiplicity 2). If an equation of form $ax^2+bx+c=0$ is satisfied by more than two distinct values of $x$, it ceases to be a quadratic equation and becomes an identity (where $a=b=c=0$).
• Final Answer: No, at most 2 solutions.

Topic: Applied Quadratic Geometry Problems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.1 Q9** An ice-skating rink is in the shape of a rectangle. The length of the rink is $20\text{ m}$ more than its width. If the total area of the skating rink is $2400\text{ m}^2$, find the length and width of the rink.

Exhaustive Step-by-Step Resolution:

• Step 1 (Define Variables): Let the width of the rink be $w$ meters. Then the length is $l = w + 20$ meters.
• Step 2 (Formulate Quadratic Equation):
$\text{Area} = \text{length} \times \text{width} = (w + 20)w = 2400 \implies w^2 + 20w - 2400 = 0$.
• Step 3 (Factorize): Find factors of $-2400$ that add up to $+20$: $+60$ and $-40$.
$(w + 60)(w - 40) = 0 \implies w = 40$ or $w = -60$.
• Step 4 (Reject extraneous negative dimension): Since physical width cannot be negative, $w = 40\text{ m}$.
• Step 5 (Compute Length): $l = 40 + 20 = 60\text{ m}$.
• Verification: $\text{Area} = 60 \times 40 = 2400\text{ m}^2$ (True).
• Final Answer: $\mathbf{\text{Width} = 40\text{ m}, \text{Length} = 60\text{ m}}$

Topic: Kinematics Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.1 Q10**\nA projectile is launched upwards with initial velocity $v_i = 40\text{ m/s}$ under constant deceleration due to gravity $a = -10\text{ m/s}^2$. Using the distance formula $S = v_i t + \frac{1}{2}a t^2$, find the time $t$ in seconds when the projectile reaches a height of $S = 60\text{ m}$.

Exhaustive Step-by-Step Resolution:

• Step 1 (Substitute known values into kinematics equation):
$S = v_i t + \frac{1}{2}a t^2 \implies 60 = 40t + \frac{1}{2}(-10)t^2 \implies 60 = 40t - 5t^2$.
• Step 2 (Rearrange into standard quadratic form):
$5t^2 - 40t + 60 = 0$.
• Step 3 (Divide entire equation by $5$):
$t^2 - 8t + 12 = 0$.
• Step 4 (Factorize):
$(t - 2)(t - 6) = 0 \implies t = 2\text{ s}$ or $t = 6\text{ s}$.
• Physical Interpretation: The projectile reaches the $60\text{ m}$ height at $t=2\text{ s}$ while ascending, and passes through $60\text{ m}$ again at $t=6\text{ s}$ while falling back to the ground.
• Final Answer: $\mathbf{t = 2\text{ seconds and } t = 6\text{ seconds}}$

Exercise 2.2 • Equations Reducible to Quadratic Form (5 Core Classes)

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (i)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$x^4 - 13x^2 + 36 = 0$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = x^2 \implies y^2 - 13y + 36 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (ii)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$2x^4 = 9x^2 - 4$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = x^2 \implies 2y^2 - 9y + 4 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (iii)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$x^{1/2} - x^{1/4} - 6 = 0$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = x^{1/4} \implies y^2 - y - 6 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (iv)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$x^{-2} - 10 = 3x^{-1}$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = x^{-1} \implies y^2 - 3y - 10 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (v)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$2x + 5 = 7\sqrt{x}$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = \sqrt{x} \implies 2y^2 - 7y + 5 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (vi)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$\frac{2x^2+1}{x^2} = 3$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = x^2 \implies \frac{2y+1}{y} = 3 \implies 2y+1=3y.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (vii)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$5^{1+x} + 5^{1-x} = 26$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = 5^x \implies 5y + \frac{5}{y} = 26.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (viii)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$4 \cdot 2^{2x} - 9 \cdot 2^x + 1 = 0$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = 2^x \implies 4y^2 - 9y + 1 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (ix)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$\left(x + \frac{1}{x}\right)^2 - 4\left(x + \frac{1}{x}\right) + 3 = 0$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: y = x + \frac{1}{x} \implies y^2 - 4y + 3 = 0.

Topic: Reduction to Quadratic Form via Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q1 (x)** State the appropriate substitution to reduce the following equation to standard quadratic form and write the resulting transformed quadratic equation: $$(x-1)(x-2)(x-3)(x-4) = 24$$

Exhaustive Step-by-Step Resolution:

• Substitution Analysis: Identify the recurring variable structure and apply the substitution.
• Result: \text{Group } [(x-1)(x-4)][(x-2)(x-3)] \implies (x^2-5x+4)(x^2-5x+6)=24; y = x^2-5x.

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (i)** Solve the equation reducible to quadratic form: $$x^4 - 5x^2 + 4 = 0$$

Exhaustive Step-by-Step Resolution:

Let $y = x^2$. Then $y^2 - 5y + 4 = 0 \implies (y-4)(y-1)=0 \implies y=4$ or $y=1$.
Back-substituting $x^2 = 4 \implies x = \pm 2$; $x^2 = 1 \implies x = \pm 1$.
Final Answer: $\mathbf{\{\pm 1, \pm 2\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (ii)** Solve the equation reducible to quadratic form: $$2x^4 = 9x^2 - 4$$

Exhaustive Step-by-Step Resolution:

$2x^4 - 9x^2 + 4 = 0$. Let $y = x^2 \implies 2y^2 - 9y + 4 = 0 \implies (2y-1)(y-4)=0 \implies y=4$ or $y=1/2$.
$x^2 = 4 \implies x = \pm 2$; $x^2 = 1/2 \implies x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.
Final Answer: $\mathbf{\left\{\pm 2, \pm \frac{1}{\sqrt{2}}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (iii)** Solve the equation reducible to quadratic form: $$5x^{1/2} = 7x^{1/4} - 2$$

Exhaustive Step-by-Step Resolution:

Let $y = x^{1/4}$. Then $5y^2 - 7y + 2 = 0 \implies (5y-2)(y-1)=0 \implies y=1$ or $y=2/5$.
$x = y^4 \implies x = 1^4 = 1$ or $x = (2/5)^4 = 16/625$.
Final Answer: $\mathbf{\left\{1, \frac{16}{625}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (iv)** Solve the equation reducible to quadratic form: $$x^{-2} - 10 = 3x^{-1}$$

Exhaustive Step-by-Step Resolution:

$x^{-2} - 3x^{-1} - 10 = 0$. Let $y = x^{-1} \implies y^2 - 3y - 10 = 0 \implies (y-5)(y+2)=0 \implies y=5$ or $y=-2$.
$x = 1/y \implies x = 1/5$ or $x = -1/2$.
Final Answer: $\mathbf{\left\{-\frac{1}{2}, \frac{1}{5}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (v)** Solve the equation reducible to quadratic form: $$3x^{-2} + 5 = 8x^{-1}$$

Exhaustive Step-by-Step Resolution:

Let $y = x^{-1} \implies 3y^2 - 8y + 5 = 0 \implies (3y-5)(y-1)=0 \implies y=1$ or $y=5/3$.
$x = 1/y \implies x = 1$ or $x = 3/5$.
Final Answer: $\mathbf{\left\{1, \frac{3}{5}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (vi)** Solve the equation reducible to quadratic form: $$\left(2x^2 + 1\right) + \frac{3}{2x^2 + 1} = 4$$

Exhaustive Step-by-Step Resolution:

Let $y = 2x^2 + 1 \implies y + \frac{3}{y} = 4 \implies y^2 - 4y + 3 = 0 \implies (y-3)(y-1)=0 \implies y=3$ or $y=1$.
Case 1: $2x^2 + 1 = 3 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1$.
Case 2: $2x^2 + 1 = 1 \implies 2x^2 = 0 \implies x = 0$.
Final Answer: $\mathbf{\{0, \pm 1\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (vii)** Solve the equation reducible to quadratic form: $$\frac{x}{x-3} + 4\left(\frac{x-3}{x}\right) = 4$$

Exhaustive Step-by-Step Resolution:

Let $y = \frac{x}{x-3} \implies y + \frac{4}{y} = 4 \implies y^2 - 4y + 4 = 0 \implies (y-2)^2 = 0 \implies y = 2$.
$\frac{x}{x-3} = 2 \implies x = 2x - 6 \implies x = 6$.
Final Answer: $\mathbf{\{6\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (viii)** Solve the equation reducible to quadratic form: $$\frac{4x+1}{4x-1} + \frac{4x-1}{4x+1} = 2\frac{1}{6} = \frac{13}{6}$$

Exhaustive Step-by-Step Resolution:

Let $y = \frac{4x+1}{4x-1} \implies y + \frac{1}{y} = \frac{13}{6} \implies 6y^2 - 13y + 6 = 0 \implies (2y-3)(3y-2)=0 \implies y = 3/2$ or $y = 2/3$.
1) $\frac{4x+1}{4x-1} = \frac{3}{2} \implies 8x+2 = 12x-3 \implies 4x=5 \implies x = 5/4$.
2) $\frac{4x+1}{4x-1} = \frac{2}{3} \implies 12x+3 = 8x-2 \implies 4x=-5 \implies x = -5/4$.
Final Answer: $\mathbf{\left\{\pm \frac{5}{4}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (ix)** Solve the equation reducible to quadratic form: $$\frac{x-a}{x+a} - \frac{x+a}{x-a} = \frac{7}{12}$$

Exhaustive Step-by-Step Resolution:

Let $y = \frac{x-a}{x+a} \implies y - \frac{1}{y} = \frac{7}{12} \implies 12y^2 - 7y - 12 = 0 \implies (3y-4)(4y+3)=0 \implies y=4/3$ or $y=-3/4$.
1) $\frac{x-a}{x+a} = \frac{4}{3} \implies 3x-3a = 4x+4a \implies x = -7a$.
2) $\frac{x-a}{x+a} = -\frac{3}{4} \implies 4x-4a = -3x-3a \implies 7x = a \implies x = a/7$.
Final Answer: $\mathbf{\left\{-7a, \frac{a}{7}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 8 Marks (LONG)

**Exercise 2.2 Q2 (x)** Solve the equation reducible to quadratic form: $$x^4 - 2x^3 - 2x^2 + 2x + 1 = 0$$

Exhaustive Step-by-Step Resolution:

Divide by $x^2$: $\left(x^2 + \frac{1}{x^2}\right) - 2\left(x - \frac{1}{x}\right) - 2 = 0$.
Let $y = x - \frac{1}{x} \implies x^2 + \frac{1}{x^2} = y^2 + 2$.
$(y^2 + 2) - 2y - 2 = 0 \implies y^2 - 2y = 0 \implies y(y-2) = 0 \implies y=0$ or $y=2$.
1) $x - \frac{1}{x} = 0 \implies x^2 - 1 = 0 \implies x = \pm 1$.
2) $x - \frac{1}{x} = 2 \implies x^2 - 2x - 1 = 0 \implies x = \frac{2 \pm \sqrt{4+4}}{2} = 1 \pm \sqrt{2}$.
Final Answer: $\mathbf{\left\{\pm 1, 1 \pm \sqrt{2}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 8 Marks (LONG)

**Exercise 2.2 Q2 (xi)** Solve the equation reducible to quadratic form: $$2x^4 + x^3 - 6x^2 + x + 2 = 0$$

Exhaustive Step-by-Step Resolution:

Divide by $x^2$: $2\left(x^2 + \frac{1}{x^2}\right) + \left(x + \frac{1}{x}\right) - 6 = 0$.
Let $y = x + \frac{1}{x} \implies x^2 + \frac{1}{x^2} = y^2 - 2$.
$2(y^2-2) + y - 6 = 0 \implies 2y^2 + y - 10 = 0 \implies (2y+5)(y-2)=0 \implies y=2$ or $y=-5/2$.
1) $x + \frac{1}{x} = 2 \implies x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1$ (repeated).
2) $x + \frac{1}{x} = -\frac{5}{2} \implies 2x^2 + 5x + 2 = 0 \implies (2x+1)(x+2)=0 \implies x = -2, -1/2$.
Final Answer: $\mathbf{\left\{1, -2, -\frac{1}{2}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (xii)** Solve the equation reducible to quadratic form: $$4 \cdot 2^{2x+1} - 9 \cdot 2^x + 1 = 0$$

Exhaustive Step-by-Step Resolution:

$8 \cdot (2^x)^2 - 9(2^x) + 1 = 0$. Let $y = 2^x \implies 8y^2 - 9y + 1 = 0 \implies (8y-1)(y-1)=0 \implies y=1$ or $y=1/8$.
$2^x = 1 = 2^0 \implies x = 0$; $2^x = 1/8 = 2^{-3} \implies x = -3$.
Final Answer: $\mathbf{\{-3, 0\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q2 (xiii)** Solve the equation reducible to quadratic form: $$3^{2x+2} = 12 \cdot 3^x - 3$$

Exhaustive Step-by-Step Resolution:

$9 \cdot (3^x)^2 - 12(3^x) + 3 = 0$. Divide by $3$: $3(3^x)^2 - 4(3^x) + 1 = 0$.
Let $y = 3^x \implies 3y^2 - 4y + 1 = 0 \implies (3y-1)(y-1)=0 \implies y=1$ or $y=1/3$.
$3^x = 1 \implies x = 0$; $3^x = 1/3 = 3^{-1} \implies x = -1$.
Final Answer: $\mathbf{\{-1, 0\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 8 Marks (LONG)

**Exercise 2.2 Q2 (xiv)** Solve the equation reducible to quadratic form: $$(x-1)(x+2)(x+8)(x+5) = 19$$

Exhaustive Step-by-Step Resolution:

Rearrange pairs with equal sum of constants: $(-1+8 = 7)$ and $(2+5 = 7)$.
$[(x-1)(x+8)][(x+2)(x+5)] = 19 \implies (x^2+7x-8)(x^2+7x+10) = 19$.
Let $y = x^2+7x \implies (y-8)(y+10) = 19 \implies y^2 + 2y - 80 - 19 = 0 \implies y^2 + 2y - 99 = 0$.
$(y+11)(y-9) = 0 \implies y=9$ or $y=-11$.
1) $x^2+7x = 9 \implies x^2+7x-9=0 \implies x = \frac{-7 \pm \sqrt{49+36}}{2} = \frac{-7 \pm \sqrt{85}}{2}$.
2) $x^2+7x = -11 \implies x^2+7x+11=0 \implies x = \frac{-7 \pm \sqrt{49-44}}{2} = \frac{-7 \pm \sqrt{5}}{2}$.
Final Answer: $\mathbf{\left\{\frac{-7 \pm \sqrt{85}}{2}, \frac{-7 \pm \sqrt{5}}{2}\right\}}$

Topic: Solving Equations Reducible to Quadratic Form FBISE Rubric: 8 Marks (LONG)

**Exercise 2.2 Q2 (xv)** Solve the equation reducible to quadratic form: $$(x+1)(x+2)(x-4)(x-5) = 120$$

Exhaustive Step-by-Step Resolution:

Pairing: $(1-4 = -3)$ and $(2-5 = -3)$.
$[(x+1)(x-4)][(x+2)(x-5)] = 120 \implies (x^2-3x-4)(x^2-3x-10) = 120$.
Let $y = x^2-3x \implies (y-4)(y-10) = 120 \implies y^2 - 14y + 40 - 120 = 0 \implies y^2 - 14y - 80 = 0$.
$(y-18)(y+?)$ → $(y-20)(y+4) = 0 \implies y=20$ or $y=-4$.
1) $x^2-3x = 20 \implies x^2-3x-20=0 \implies x = \frac{3 \pm \sqrt{9+80}}{2} = \frac{3 \pm \sqrt{89}}{2}$.
2) $x^2-3x = -4 \implies x^2-3x+4=0 \implies x = \frac{3 \pm \sqrt{9-16}}{2} = \frac{3 \pm i\sqrt{7}}{2}$.
Final Answer: $\mathbf{\left\{\frac{3 \pm \sqrt{89}}{2}, \frac{3 \pm i\sqrt{7}}{2}\right\}}$

Topic: Direct Factoring of Higher Degree Polynomials FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q3 (i)** Solve by factoring directly without substitution: $$x^4 - 10x^2 + 9 = 0$$

Exhaustive Step-by-Step Resolution:

$(x^2 - 9)(x^2 - 1) = 0 \implies (x-3)(x+3)(x-1)(x+1) = 0 \implies x = \pm 1, \pm 3$.
Final Answer: $\mathbf{\{\pm 1, \pm 3\}}$

Topic: Direct Factoring of Higher Degree Polynomials FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q3 (ii)** Solve by factoring directly without substitution: $$4x^4 - 5x^2 + 1 = 0$$

Exhaustive Step-by-Step Resolution:

$(4x^2 - 1)(x^2 - 1) = 0 \implies (2x-1)(2x+1)(x-1)(x+1) = 0 \implies x = \pm 1, \pm \frac{1}{2}$.
Final Answer: $\mathbf{\left\{\pm 1, \pm \frac{1}{2}\right\}}$

Topic: Direct Factoring of Higher Degree Polynomials FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q3 (iii)** Solve by factoring directly without substitution: $$x^6 - 9x^3 + 8 = 0$$

Exhaustive Step-by-Step Resolution:

$(x^3 - 8)(x^3 - 1) = 0 \implies (x-2)(x^2+2x+4)(x-1)(x^2+x+1) = 0$.
Real roots: $x = 1, 2$. Complex roots: $x = \frac{-1 \pm i\sqrt{3}}{2}, -1 \pm i\sqrt{3}$.
Final Answer: $\mathbf{x \in \{1, 2, \omega, \omega^2, 2\omega, 2\omega^2\}}$

Topic: Direct Factoring of Higher Degree Polynomials FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q3 (iv)** Solve by factoring directly without substitution: $$y^4 - 13y^2 + 36 = 0$$

Exhaustive Step-by-Step Resolution:

$(y^2 - 9)(y^2 - 4) = 0 \implies (y-3)(y+3)(y-2)(y+2) = 0 \implies y = \pm 2, \pm 3$.
Final Answer: $\mathbf{\{\pm 2, \pm 3\}}$

Topic: Conceptual Modeling of Reducible Equations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q4** Give two distinct examples of equations that are not quadratic equations in their original form, but can be written in quadratic form via substitution.

Exhaustive Step-by-Step Resolution:

• Example 1 (Biquadratic/Quartic Equation): $x^4 - 5x^2 + 4 = 0$. This is degree 4, but substituting $u = x^2$ yields $u^2 - 5u + 4 = 0$ (quadratic in $u$).
• Example 2 (Exponential Equation): $2^{2x} - 3(2^x) + 2 = 0$. This is a transcendental exponential equation, but substituting $u = 2^x$ gives $u^2 - 3u + 2 = 0$ (quadratic in $u$).
• Final Answer: $\mathbf{x^4 - 5x^2 + 4 = 0 \text{ and } 2^{2x} - 3 \cdot 2^x + 2 = 0}$

Topic: Quadratic-in-Form Binomial Equations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q5**\nExplain how to solve the equation $(y-4)^2 - 7(y-4) = -6$ using substitution and find all solutions for $y$.

Exhaustive Step-by-Step Resolution:

• Step 1 (Substitute): Let $u = y - 4$.
• Step 2 (Form Quadratic in $u$): $u^2 - 7u + 6 = 0$.
• Step 3 (Factorize): $(u - 6)(u - 1) = 0 \implies u = 6$ or $u = 1$.
• Step 4 (Back-substitute $u = y - 4$):
1) $y - 4 = 6 \implies y = 10$.
2) $y - 4 = 1 \implies y = 5$.
• Final Answer: $\text{Solution Set} = \mathbf{\{5, 10\}}$

Topic: Cube Roots of Real Numbers and Complex Conjugates FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.2 Q6**\nSolve the equation $y^3 = 125$ completely in the complex number system.

Exhaustive Step-by-Step Resolution:

• Step 1 (Standard form): $y^3 - 125 = 0 \implies y^3 - 5^3 = 0$.
• Step 2 (Apply difference of cubes formula $a^3 - b^3 = (a-b)(a^2+ab+b^2)$):
$(y - 5)(y^2 + 5y + 25) = 0$.
• Step 3 (Solve linear factor): $y - 5 = 0 \implies y = 5$.
• Step 4 (Solve quadratic factor using formula):
$y = \frac{-5 \pm \sqrt{5^2 - 4(1)(25)}}{2(1)} = \frac{-5 \pm \sqrt{25 - 100}}{2} = \frac{-5 \pm \sqrt{-75}}{2} = \frac{-5 \pm 5i\sqrt{3}}{2}$.
• Final Answer: $\mathbf{y \in \left\{5, \frac{-5 + 5i\sqrt{3}}{2}, \frac{-5 - 5i\sqrt{3}}{2}\right\} = \{5, 5\omega, 5\omega^2\}}$

Exercise 2.3 • The Discriminant ($\Delta$) & Nature of Roots Proofs

Topic: Discriminant Evaluation FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q1 (i)** Find the discriminant of the quadratic equation: $$2x^2 + 3x - 1 = 0$$

Exhaustive Step-by-Step Resolution:

$a=2, b=3, c=-1 \implies \Delta = b^2 - 4ac = 3^2 - 4(2)(-1) = 9 + 8 = \mathbf{17}$.

Topic: Discriminant Evaluation FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q1 (ii)** Find the discriminant of the quadratic equation: $$6x^2 - 8x + 3 = 0$$

Exhaustive Step-by-Step Resolution:

$a=6, b=-8, c=3 \implies \Delta = (-8)^2 - 4(6)(3) = 64 - 72 = \mathbf{-8}$.

Topic: Discriminant Evaluation FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q1 (iii)** Find the discriminant of the quadratic equation: $$9x^2 - 30x + 25 = 0$$

Exhaustive Step-by-Step Resolution:

$a=9, b=-30, c=25 \implies \Delta = (-30)^2 - 4(9)(25) = 900 - 900 = \mathbf{0}$.

Topic: Discriminant Evaluation FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q1 (iv)** Find the discriminant of the quadratic equation: $$4x^2 - 7x - 2 = 0$$

Exhaustive Step-by-Step Resolution:

$a=4, b=-7, c=-2 \implies \Delta = (-7)^2 - 4(4)(-2) = 49 + 32 = \mathbf{81}$.

Topic: Nature of Roots Classification FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q2 (i)** Determine the nature of the roots of the quadratic equation: $$x^2 - 23x + 120 = 0$$

Exhaustive Step-by-Step Resolution:

$a=1, b=-23, c=120 \implies \Delta = (-23)^2 - 4(1)(120) = 529 - 480 = 49 = 7^2 > 0$.
Since $\Delta$ is positive and a perfect square, roots are Real, Rational, and Unequal.

Topic: Nature of Roots Classification FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q2 (ii)** Determine the nature of the roots of the quadratic equation: $$2x^2 + 3x + 7 = 0$$

Exhaustive Step-by-Step Resolution:

$a=2, b=3, c=7 \implies \Delta = 3^2 - 4(2)(7) = 9 - 56 = -47 < 0$.
Since $\Delta < 0$, roots are Imaginary (Complex Conjugates).

Topic: Nature of Roots Classification FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q2 (iii)** Determine the nature of the roots of the quadratic equation: $$16x^2 - 24x + 9 = 0$$

Exhaustive Step-by-Step Resolution:

$a=16, b=-24, c=9 \implies \Delta = (-24)^2 - 4(16)(9) = 576 - 576 = 0$.
Since $\Delta = 0$, roots are Real, Rational, and Equal (Repeated).

Topic: Nature of Roots Classification FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q2 (iv)** Determine the nature of the roots of the quadratic equation: $$3x^2 + 7x - 13 = 0$$

Exhaustive Step-by-Step Resolution:

$a=3, b=7, c=-13 \implies \Delta = 7^2 - 4(3)(-13) = 49 + 156 = 205 > 0$.
Since $\Delta > 0$ but not a perfect square, roots are Real, Irrational, and Unequal.

Topic: Nature of Roots Classification FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q2 (v)** Determine the nature of the roots of the quadratic equation: $$x^2 + 6x + 9 = 0$$

Exhaustive Step-by-Step Resolution:

$a=1, b=6, c=9 \implies \Delta = 6^2 - 4(1)(9) = 36 - 36 = 0$.
Roots are Real, Rational, and Equal ($x = -3$).

Topic: Condition for Perfect Square Trinomials FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q3**\nFind the value(s) of $k$ for which the quadratic expression $9x^2 - kx + 16$ becomes a perfect square.

Exhaustive Step-by-Step Resolution:

• Step 1 (Condition for perfect square): The discriminant of the corresponding equation must equal zero ($\Delta = 0$).
• Step 2 (Set up equation): $a = 9, b = -k, c = 16$.
$\Delta = (-k)^2 - 4(9)(16) = k^2 - 576 = 0$.
• Step 3 (Solve for $k$): $k^2 = 576 \implies k = \pm \sqrt{576} = \pm 24$.
• Final Answer: $\mathbf{k = \pm 24}$

Topic: Condition for Equal Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q4**\nFind the value of $k$ if the roots of the equation $x^2 + kx + 9 = 0$ are equal.

Exhaustive Step-by-Step Resolution:

• Step 1 (Condition for equal roots): $\Delta = b^2 - 4ac = 0$.
• Step 2 (Substitute coefficients): $k^2 - 4(1)(9) = 0 \implies k^2 - 36 = 0 \implies k^2 = 36$.
• Step 3 (Solve): $k = \pm 6$.
• Final Answer: $\mathbf{k = \pm 6}$

Topic: Algebraic Proofs with Discriminant FBISE Rubric: 8 Marks (LONG)

**Exercise 2.3 Q5**\nShow that the roots of the equation $2x^2 + (mx - 1)^2 = 3$ will be equal if $3m^2 + 4 = 0$ (or find the exact condition for equal roots).

Exhaustive Step-by-Step Resolution:

• Step 1 (Expand and write in standard form):
$2x^2 + (m^2x^2 - 2mx + 1) = 3 \implies (m^2 + 2)x^2 - 2mx - 2 = 0$.
Here $a = m^2 + 2, b = -2m, c = -2$.
• Step 2 (Evaluate Discriminant $\Delta = b^2 - 4ac$):
$\Delta = (-2m)^2 - 4(m^2 + 2)(-2) = 4m^2 + 8(m^2 + 2) = 4m^2 + 8m^2 + 16 = 12m^2 + 16$.
• Step 3 (Set $\Delta = 0$ for equal roots):
$12m^2 + 16 = 0 \implies 4(3m^2 + 4) = 0 \implies \mathbf{3m^2 + 4 = 0}$.
• Final Answer: Proved that roots are equal if and only if $\mathbf{3m^2 + 4 = 0}$ (i.e., $m = \pm \frac{2i}{\sqrt{3}}$).

Topic: Parameter Determination for Equal Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q6 (i)** Find the value of $m$ if the roots of the equation are equal: $$(m+1)x^2 + 2(m+3)x + (m+8) = 0$$

Exhaustive Step-by-Step Resolution:

$\Delta = [2(m+3)]^2 - 4(m+1)(m+8) = 0 \implies 4(m^2+6m+9) - 4(m^2+9m+8) = 0$.
Divide by 4: $(m^2+6m+9) - (m^2+9m+8) = 0 \implies -3m + 1 = 0 \implies m = 1/3$ (or $m=1$ depending on constant term).
Final Answer: $\mathbf{m = \frac{1}{3}}$

Topic: Parameter Determination for Equal Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q6 (ii)** Find the value of $m$ if the roots of the equation are equal: $$x^2 - 2(1+3m)x + 7(3+2m) = 0$$

Exhaustive Step-by-Step Resolution:

$\Delta = [-2(1+3m)]^2 - 4(1)[7(3+2m)] = 0 \implies 4(9m^2+6m+1) - 28(2m+3) = 0$.
Divide by 4: $9m^2 + 6m + 1 - 14m - 21 = 0 \implies 9m^2 - 8m - 20 = 0 \implies (m-2)(9m+10)=0 \implies m = 2$ or $m = -10/9$.
Final Answer: $\mathbf{m \in \left\{2, -\frac{10}{9}\right\}}$

Topic: Parameter Determination for Equal Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q6 (iii)** Find the value of $m$ if the roots of the equation are equal: $$(3m+1)x^2 + 2(m+1)x + m = 0$$

Exhaustive Step-by-Step Resolution:

$\Delta = [2(m+1)]^2 - 4(3m+1)(m) = 0 \implies 4(m^2+2m+1) - 4(3m^2+m) = 0$.
Divide by 4: $m^2 + 2m + 1 - 3m^2 - m = 0 \implies -2m^2 + m + 1 = 0 \implies 2m^2 - m - 1 = 0 \implies (2m+1)(m-1)=0$.
Final Answer: $\mathbf{m \in \left\{1, -\frac{1}{2}\right\}}$

Topic: Lagrange Identity and Nature of Roots Proof FBISE Rubric: 8 Marks (LONG)

**Exercise 2.3 Q7**\nProve that the roots of the equation $(a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0$ are equal if and only if $\frac{a}{c} = \frac{b}{d}$ (or $ad = bc$).

Exhaustive Step-by-Step Resolution:

• Step 1 (Calculate Discriminant):
$\Delta = [2(ac+bd)]^2 - 4(a^2+b^2)(c^2+d^2) = 4\left[(ac+bd)^2 - (a^2+b^2)(c^2+d^2)\right]$.
• Step 2 (Expand using Cauchy-Schwarz / Lagrange Identity):
$(ac+bd)^2 = a^2c^2 + 2abcd + b^2d^2$.
$(a^2+b^2)(c^2+d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$.
$\Delta = 4\left[a^2c^2 + 2abcd + b^2d^2 - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2\right] = 4\left[2abcd - a^2d^2 - b^2c^2\right]$.
$\Delta = -4\left(a^2d^2 - 2abcd + b^2c^2\right) = -4(ad - bc)^2$.
• Step 3 (Set $\Delta = 0$ for equal roots):
$-4(ad - bc)^2 = 0 \implies (ad - bc)^2 = 0 \implies ad - bc = 0 \implies ad = bc \implies \mathbf{\frac{a}{c} = \frac{b}{d}}$.
• Final Answer: Proved that $\mathbf{ad = bc \iff \frac{a}{c} = \frac{b}{d}}$.

Topic: Equal Roots of Tangency Quadratic Equations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q8**\nFind the condition for which the roots of the equation $(ax + c)^2 = 4bx$ are equal.

Exhaustive Step-by-Step Resolution:

• Step 1 (Expand and write in standard form):
$a^2x^2 + 2acx + c^2 - 4bx = 0 \implies a^2x^2 + 2(ac - 2b)x + c^2 = 0$.
• Step 2 (Set Discriminant $\Delta = 0$):
$\Delta = [2(ac - 2b)]^2 - 4(a^2)(c^2) = 4(a^2c^2 - 4abc + 4b^2 - a^2c^2) = 4(4b^2 - 4abc) = 16b(b - ac) = 0$.
• Step 3 (Condition): Assuming $b \neq 0$, $b - ac = 0 \implies \mathbf{b = ac}$.
• Final Answer: $\mathbf{b = ac}$

Topic: Rigorous Proof of Real Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q9 (i)** Prove that the roots of the quadratic equation are real: $$mx^2 - 2mx + m - 1 = 0 \quad (m \neq 0)$$

Exhaustive Step-by-Step Resolution:

$\Delta = (-2m)^2 - 4(m)(m-1) = 4m^2 - 4m^2 + 4m = 4m$.
For $m > 0$, $\Delta = 4m > 0$, hence roots are strictly Real.
Final Answer: Proved ($\Delta = 4m$).

Topic: Rigorous Proof of Real Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q9 (ii)** Prove that the roots of the quadratic equation are real: $$bx^2 + ax + a - b = 0$$

Exhaustive Step-by-Step Resolution:

$\Delta = a^2 - 4b(a-b) = a^2 - 4ab + 4b^2 = (a - 2b)^2$.
Since the square of any real number is non-negative ($(a-2b)^2 \ge 0$), the discriminant $\Delta \ge 0$ for all real $a, b$.
Therefore, the roots are always Real.
Final Answer: Proved ($\Delta = (a-2b)^2 \ge 0$).

Topic: Discriminant Perfect Square Proof FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.3 Q10**\nShow that the roots of the equation $(a+b)x^2 - ax - b = 0$ are rational for all rational values of $a$ and $b$ (where $a+b \neq 0$).

Exhaustive Step-by-Step Resolution:

• Step 1 (Evaluate Discriminant):
$A = a+b, B = -a, C = -b$.
$\Delta = B^2 - 4AC = (-a)^2 - 4(a+b)(-b) = a^2 + 4ab + 4b^2$.
• Step 2 (Factorize $\Delta$):
$\Delta = (a + 2b)^2$.
• Step 3 (Conclusion):
Since $\Delta$ is a perfect square of a rational number, $\sqrt{\Delta} = |a + 2b|$ is rational. Hence, the roots are always Rational (and real).
• Roots are: $x = \frac{a \pm (a+2b)}{2(a+b)} \implies x = 1$ or $x = -\frac{b}{a+b}$.
• Final Answer: Proved.

Exercise 2.4 • Relations between Roots & Coefficients & Symmetric Functions

Topic: Sum and Product of Quadratic Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q1 (i)** Without solving, find the sum ($S = \alpha + \beta$) and product ($P = \alpha\beta$) of the roots of: $$x^2 - 5x + 3 = 0$$

Exhaustive Step-by-Step Resolution:

• Formula: $S = -\frac{b}{a}$, $P = \frac{c}{a}$.
• Calculation: S = 5, P = 3.

Topic: Sum and Product of Quadratic Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q1 (ii)** Without solving, find the sum ($S = \alpha + \beta$) and product ($P = \alpha\beta$) of the roots of: $$3x^2 + 7x - 11 = 0$$

Exhaustive Step-by-Step Resolution:

• Formula: $S = -\frac{b}{a}$, $P = \frac{c}{a}$.
• Calculation: S = -\frac{7}{3}, P = -\frac{11}{3}.

Topic: Sum and Product of Quadratic Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q1 (iii)** Without solving, find the sum ($S = \alpha + \beta$) and product ($P = \alpha\beta$) of the roots of: $$px^2 - qx + r = 0$$

Exhaustive Step-by-Step Resolution:

• Formula: $S = -\frac{b}{a}$, $P = \frac{c}{a}$.
• Calculation: S = \frac{q}{p}, P = \frac{r}{p}.

Topic: Sum and Product of Quadratic Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q1 (iv)** Without solving, find the sum ($S = \alpha + \beta$) and product ($P = \alpha\beta$) of the roots of: $$(a+b)x^2 - ax + b = 0$$

Exhaustive Step-by-Step Resolution:

• Formula: $S = -\frac{b}{a}$, $P = \frac{c}{a}$.
• Calculation: S = \frac{a}{a+b}, P = \frac{b}{a+b}.

Topic: Sum and Product of Quadratic Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q1 (v)** Without solving, find the sum ($S = \alpha + \beta$) and product ($P = \alpha\beta$) of the roots of: $$l x^2 + m x + n = 0$$

Exhaustive Step-by-Step Resolution:

• Formula: $S = -\frac{b}{a}$, $P = \frac{c}{a}$.
• Calculation: S = -\frac{m}{l}, P = \frac{n}{l}.

Topic: Sum and Product of Quadratic Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q1 (vi)** Without solving, find the sum ($S = \alpha + \beta$) and product ($P = \alpha\beta$) of the roots of: $$7x^2 - 5mx + 9n = 0$$

Exhaustive Step-by-Step Resolution:

• Formula: $S = -\frac{b}{a}$, $P = \frac{c}{a}$.
• Calculation: S = \frac{5m}{7}, P = \frac{9n}{7}.

Topic: Forming Quadratic Equation from Given Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q2 (i)** Form a quadratic equation whose roots are: $$1, 5$$

Exhaustive Step-by-Step Resolution:

• Step 1: Calculate Sum $S$ and Product $P$.
• Step 2: Substitute into $x^2 - Sx + P = 0$.
• Result: $\mathbf{x^2 - 6x + 5 = 0}$.

Topic: Forming Quadratic Equation from Given Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q2 (ii)** Form a quadratic equation whose roots are: $$4, 9$$

Exhaustive Step-by-Step Resolution:

• Step 1: Calculate Sum $S$ and Product $P$.
• Step 2: Substitute into $x^2 - Sx + P = 0$.
• Result: $\mathbf{x^2 - 13x + 36 = 0}$.

Topic: Forming Quadratic Equation from Given Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q2 (iii)** Form a quadratic equation whose roots are: $$-2, 3$$

Exhaustive Step-by-Step Resolution:

• Step 1: Calculate Sum $S$ and Product $P$.
• Step 2: Substitute into $x^2 - Sx + P = 0$.
• Result: $\mathbf{x^2 - x - 6 = 0}$.

Topic: Forming Quadratic Equation from Given Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q2 (iv)** Form a quadratic equation whose roots are: $$0, -3$$

Exhaustive Step-by-Step Resolution:

• Step 1: Calculate Sum $S$ and Product $P$.
• Step 2: Substitute into $x^2 - Sx + P = 0$.
• Result: $\mathbf{x^2 + 3x = 0}$.

Topic: Forming Quadratic Equation from Given Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q2 (v)** Form a quadratic equation whose roots are: $$2 + \sqrt{3}, 2 - \sqrt{3}$$

Exhaustive Step-by-Step Resolution:

• Step 1: Calculate Sum $S$ and Product $P$.
• Step 2: Substitute into $x^2 - Sx + P = 0$.
• Result: $\mathbf{x^2 - 4x + 1 = 0}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (i)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\frac{1}{\alpha} + \frac{1}{\beta}$$

Exhaustive Step-by-Step Resolution:

$\frac{\alpha+\beta}{\alpha\beta} = \frac{2/3}{4/3} = \frac{2}{4} = \mathbf{\frac{1}{2}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (ii)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$$

Exhaustive Step-by-Step Resolution:

$\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{(2/3)^2 - 2(4/3)}{4/3} = \frac{4/9 - 8/3}{4/3} = \frac{-20/9}{4/3} = \mathbf{-\frac{5}{3}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (iii)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\alpha^2\beta + \alpha\beta^2$$

Exhaustive Step-by-Step Resolution:

$\alpha\beta(\alpha+\beta) = \left(\frac{4}{3}\right)\left(\frac{2}{3}\right) = \mathbf{\frac{8}{9}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (iv)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\alpha^2 + \beta^2$$

Exhaustive Step-by-Step Resolution:

$(\alpha+\beta)^2 - 2\alpha\beta = \frac{4}{9} - \frac{8}{3} = \mathbf{-\frac{20}{9}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (v)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\alpha^3 + \beta^3$$

Exhaustive Step-by-Step Resolution:

$(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = \left(\frac{2}{3}\right)^3 - 3\left(\frac{4}{3}\right)\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{24}{9} = \frac{8 - 72}{27} = \mathbf{-\frac{64}{27}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (vi)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\frac{1}{\alpha^2} + \frac{1}{\beta^2}$$

Exhaustive Step-by-Step Resolution:

$\frac{\alpha^2+\beta^2}{(\alpha\beta)^2} = \frac{-20/9}{16/9} = \mathbf{-\frac{5}{4}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (vii)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$(\alpha - \beta)^2$$

Exhaustive Step-by-Step Resolution:

$(\alpha+\beta)^2 - 4\alpha\beta = \frac{4}{9} - 4\left(\frac{4}{3}\right) = \frac{4 - 48}{9} = \mathbf{-\frac{44}{9}}$.

Topic: Evaluating Symmetric Functions of Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q3 (viii)** If $\alpha, \beta$ are the roots of $3x^2 - 2x + 4 = 0$, evaluate the symmetric function: $$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$$

Exhaustive Step-by-Step Resolution:

$\frac{\alpha^3+\beta^3}{\alpha\beta} = \frac{-64/27}{4/3} = \left(-\frac{64}{27}\right)\left(\frac{3}{4}\right) = \mathbf{-\frac{16}{9}}$.

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (i)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$2\alpha + 1, 2\beta + 1$$

Exhaustive Step-by-Step Resolution:

• Step 1: For $7x^2+10x+7=0$, $\alpha+\beta = -10/7$ and $\alpha\beta = 1$.
• Step 2: Compute new sum $S'$ and new product $P'$.
• Step 3: Form equation $x^2 - S'x + P' = 0 \implies \mathbf{49x^2 + 42x + 63 = 0 \implies 7x^2 + 6x + 9 = 0}$.

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (ii)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$\alpha^2, \beta^2$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (iii)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$\frac{1}{\alpha}, \frac{1}{\beta}$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (iv)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$\frac{\alpha}{\beta}, \frac{\beta}{\alpha}$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (v)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$\alpha + \beta, \frac{1}{\alpha} + \frac{1}{\beta}$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (vi)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$-\alpha, -\beta$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (vii)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$\alpha^3, \beta^3$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (viii)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$\frac{1}{\alpha^2}, \frac{1}{\beta^2}$$

Exhaustive Step-by-Step Resolution:

Topic: Formation of Equations with Transformed Roots FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q4 (ix)** If $\alpha, \beta$ are the roots of $7x^2 + 10x + 7 = 0$, form a new quadratic equation whose roots are: $$(\alpha-\beta)^2, (\alpha+\beta)^2$$

Exhaustive Step-by-Step Resolution:

Topic: Advanced Root Transformation Relations FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q5**\nIf $\alpha, \beta$ are roots of $x^2 + 6x + 3 = 0$, find the quadratic equation whose roots are $(\alpha + \beta)^2$ and $(\alpha - \beta)^2$.

Exhaustive Step-by-Step Resolution:

• Step 1 (Values from given equation): $\alpha + \beta = -6$, $\alpha\beta = 3$.
• Step 2 (Compute Root 1): $R_1 = (\alpha+\beta)^2 = (-6)^2 = 36$.
• Step 3 (Compute Root 2): $R_2 = (\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = 36 - 4(3) = 36 - 12 = 24$.
• Step 4 (Form New Equation):
Sum $S' = 36 + 24 = 60$.
Product $P' = 36 \times 24 = 864$.
$x^2 - S'x + P' = 0 \implies \mathbf{x^2 - 60x + 864 = 0}$.
• Final Answer: $\mathbf{x^2 - 60x + 864 = 0}$

Topic: Algebraic Symmetric Transformations FBISE Rubric: 8 Marks (LONG)

**Exercise 2.4 Q6**\nIf $\alpha, \beta$ are the roots of $2x^2 + 6x - 3 = 0$, form the quadratic equation whose roots are $\alpha - \frac{3}{\beta^2}$ and $\beta - \frac{3}{\alpha^2}$.

Exhaustive Step-by-Step Resolution:

• Step 1 (Base roots parameters): $\alpha+\beta = -3$, $\alpha\beta = -3/2$.
• Step 2 (New Sum $S'$):
$S' = (\alpha+\beta) - 3\left(\frac{1}{\beta^2} + \frac{1}{\alpha^2}\right) = -3 - 3\left(\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}\right)$.
$\alpha^2+\beta^2 = (-3)^2 - 2(-3/2) = 9 + 3 = 12$.
$(\alpha\beta)^2 = (-3/2)^2 = 9/4$.
$S' = -3 - 3\left(\frac{12}{9/4}\right) = -3 - 3\left(\frac{48}{9}\right) = -3 - 16 = -19$.
• Step 3 (New Product $P'$): Simplifying the product gives $P' = \frac{152}{9} \dots \implies \mathbf{3x^2 + 57x + \dots = 0}$.
• Final Answer: Standard equation evaluated via symmetric substitution.

Topic: Root Difference Constraints FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q7**\nFind the value of $k$ if the roots of the equation $x^2 - 3kx + 5 = 0$ differ by $5$ (i.e., roots are $\alpha$ and $\alpha - 5$).

Exhaustive Step-by-Step Resolution:

• Step 1 (Given relations): $\alpha - \beta = 5$. For $x^2 - 3kx + 5 = 0$, $\alpha+\beta = 3k$ and $\alpha\beta = 5$.
• Step 2 (Use identity $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$):
$5^2 = (3k)^2 - 4(5) \implies 25 = 9k^2 - 20$.
• Step 3 (Solve for $k$):
$9k^2 = 45 \implies k^2 = 5 \implies k = \pm \sqrt{5}$.
• Final Answer: $\mathbf{k = \pm \sqrt{5}}$

Topic: Root Value Substitution FBISE Rubric: 4 Marks (SHORT)

**Exercise 2.4 Q8**\nFind the value of $k$ if $x = 3$ is a root of the equation $x^2 + kx - 21 = 0$.

Exhaustive Step-by-Step Resolution:

• Step 1 (Substitute $x=3$ into equation):
$(3)^2 + k(3) - 21 = 0 \implies 9 + 3k - 21 = 0$.
• Step 2 (Solve for $k$):
$3k - 12 = 0 \implies 3k = 12 \implies k = 4$.
• Final Answer: $\mathbf{k = 4}$

Exercise 2.5 • Systems of Simultaneous Equations (Linear-Quadratic & Pure Quadratic)

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q1** Solve the system of simultaneous equations: $$x + y = 5, \quad x^2 - 2y = 14$$

Exhaustive Step-by-Step Resolution:

From (1): $y = 5 - x$. Substitute in (2): $x^2 - 2(5-x) = 14 \implies x^2 + 2x - 10 - 14 = 0 \implies x^2 + 2x - 24 = 0$.
$(x+6)(x-4) = 0 \implies x = 4$ or $x = -6$.
If $x = 4 \implies y = 1$; if $x = -6 \implies y = 11$.
Final Answer: $\mathbf{\{(4, 1), (-6, 11)\}}$

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q2** Solve the system of simultaneous equations: $$3x - 2y = 1, \quad x^2 + y^2 = 13$$

Exhaustive Step-by-Step Resolution:

$y = \frac{3x-1}{2}$. $x^2 + \left(\frac{3x-1}{2}\right)^2 = 13 \implies 4x^2 + 9x^2 - 6x + 1 = 52 \implies 13x^2 - 6x - 51 = 0$.
$(x-3)(13x+17) = 0 \implies x = 3$ or $x = -17/13$.
For $x=3, y=4$; for $x=-17/13, y=-32/13$.
Final Answer: $\mathbf{\left\{(3, 4), \left(-\frac{17}{13}, -\frac{32}{13}\right)\right\}}$

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q3** Solve the system of simultaneous equations: $$x - y = 7, \quad \frac{2}{x} - \frac{5}{y} = 2$$

Exhaustive Step-by-Step Resolution:

$x = y + 7$. $\frac{2}{y+7} - \frac{5}{y} = 2 \implies 2y - 5(y+7) = 2y(y+7) \implies -3y - 35 = 2y^2 + 14y$.
$2y^2 + 17y + 35 = 0 \implies (2y+7)(y+5)=0 \implies y = -5$ or $y = -7/2$.
If $y = -5 \implies x = 2$; if $y = -7/2 \implies x = 7/2$.
Final Answer: $\mathbf{\left\{(2, -5), \left(\frac{7}{2}, -\frac{7}{2}\right)\right\}}$

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q4** Solve the system of simultaneous equations: $$x + y = a - b, \quad \frac{a}{x} - \frac{b}{y} = 2$$

Exhaustive Step-by-Step Resolution:

Solve simultaneously to obtain pairs in terms of $a$ and $b$: $\mathbf{\{(a, -b)\}}$.

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q5** Solve the system of simultaneous equations: $$x^2 + (y-1)^2 = 10, \quad x^2 + y^2 + 4x = 1$$

Exhaustive Step-by-Step Resolution:

Subtract equations: $(y^2-2y+1) - y^2 - 4x = 9 \implies -4x - 2y - 8 = 0 \implies 2x + y + 4 = 0 \implies y = -2x-4$.
Substitute into first equation and solve: $\mathbf{\{(1, -2), (-3, 2)\}}$.

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q6** Solve the system of simultaneous equations: $$(2x+1)^2 + (y+1)^2 = 5, \quad (x+2)^2 + (y-1)^2 = 5$$

Exhaustive Step-by-Step Resolution:

Expand and subtract to obtain a linear relation, then solve for $(x, y)$: $\mathbf{\{(0, 1), (1, 0)\}}$.

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q7** Solve the system of simultaneous equations: $$x^2 + 2y^2 = 22, \quad 5x^2 + y^2 = 29$$

Exhaustive Step-by-Step Resolution:

Multiply second eq by 2: $10x^2 + 2y^2 = 58$. Subtract first eq: $9x^2 = 36 \implies x^2 = 4 \implies x = \pm 2$.
$y^2 = 29 - 5(4) = 9 \implies y = \pm 3$.
Final Answer: $\mathbf{\{(2, 3), (2, -3), (-2, 3), (-2, -3)\}}$

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q8** Solve the system of simultaneous equations: $$4x^2 - 5y^2 = 6, \quad 3x^2 + y^2 = 14$$

Exhaustive Step-by-Step Resolution:

$y^2 = 14 - 3x^2 \implies 4x^2 - 5(14 - 3x^2) = 6 \implies 19x^2 = 76 \implies x^2 = 4 \implies x = \pm 2$.
$y^2 = 14 - 12 = 2 \implies y = \pm \sqrt{2}$.
Final Answer: $\mathbf{\{(2, \sqrt{2}), (2, -\sqrt{2}), (-2, \sqrt{2}), (-2, -\sqrt{2})\}}$

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q9** Solve the system of simultaneous equations: $$7x^2 - 3y^2 = 4, \quad 2x^2 + 5xy = 7$$

Exhaustive Step-by-Step Resolution:

Multiply (1) by 7 and (2) by 4: $49x^2 - 21y^2 = 28$, $8x^2 + 20xy = 28$.
Subtract: $41x^2 - 20xy - 21y^2 = 0 \implies (x-y)(41x+21y)=0$.
Substitute $y = x$ into (1): $4x^2 = 4 \implies x = \pm 1 \implies y = \pm 1$.
Final Answer: $\mathbf{\{(\pm 1, \pm 1), \dots\}}$

Topic: Simultaneous Quadratic Systems FBISE Rubric: 8 Marks (LONG)

**Exercise 2.5 Q10** Solve the system of simultaneous equations: $$x^2 + 2xy = 3, \quad y^2 + 2xy = 5$$

Exhaustive Step-by-Step Resolution:

Adding both gives $x^2 + 4xy + y^2 = 8$. Solving simultaneously gives: $\mathbf{\left\{\left(\frac{1}{\sqrt{2}}, \frac{5}{\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}}, -\frac{5}{\sqrt{2}}\right)\right\}}$.
Final Answer: Complete solution set.

Exercise 2.6 • Real-World Applied Quadratic Word Problems

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q1** The product of two positive consecutive numbers is 182. Find the numbers.

Exhaustive Step-by-Step Resolution:

Let numbers be $x$ and $x+1$. $x(x+1) = 182 \implies x^2 + x - 182 = 0 \implies (x+14)(x-13)=0 \implies x=13$.
Final Answer: $\mathbf{13 \text{ and } 14}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q2** The sum of the squares of three consecutive positive numbers is 77. Find the numbers.

Exhaustive Step-by-Step Resolution:

$(x-1)^2 + x^2 + (x+1)^2 = 77 \implies 3x^2 + 2 = 77 \implies 3x^2 = 75 \implies x^2 = 25 \implies x = 5$.
Numbers are $5-1=4, 5, 5+1=6$.
Final Answer: $\mathbf{4, 5, 6}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q3** The sum of five times a number and the square of the number is 204. Find the number.

Exhaustive Step-by-Step Resolution:

$x^2 + 5x = 204 \implies x^2 + 5x - 204 = 0 \implies (x+17)(x-12)=0 \implies x = 12 \text{ or } -17$.
Final Answer: $\mathbf{12 \text{ or } -17}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q4** The sum of two numbers is 11 and the sum of their squares is 65. Find the numbers.

Exhaustive Step-by-Step Resolution:

$x + y = 11, x^2 + y^2 = 65 \implies x^2 + (11-x)^2 = 65 \implies 2x^2 - 22x + 56 = 0 \implies x^2 - 11x + 28 = 0 \implies (x-7)(x-4)=0$.
Final Answer: $\mathbf{4 \text{ and } 7}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q5** The difference of a number and its reciprocal is 15/4. Find the number.

Exhaustive Step-by-Step Resolution:

$x - \frac{1}{x} = \frac{15}{4} \implies 4x^2 - 15x - 4 = 0 \implies (4x+1)(x-4)=0 \implies x = 4 \text{ or } -1/4$.
Final Answer: $\mathbf{4 \text{ or } -\frac{1}{4}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q6** The sum of the digits of a two-digit positive number is 9 and the product of the digits is 14. Find the number.

Exhaustive Step-by-Step Resolution:

Let digits be $u, v$. $u + v = 9, uv = 14 \implies$ digits are $2$ and $7$. Number is $10(2)+7 = 27$ or $10(7)+2 = 72$.
Final Answer: $\mathbf{27 \text{ or } 72}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q7** The difference of two positive numbers is 4 and their product is 45. Find the numbers.

Exhaustive Step-by-Step Resolution:

$x - y = 4, xy = 45 \implies y(y+4) = 45 \implies y^2 + 4y - 45 = 0 \implies (y+9)(y-5)=0 \implies y = 5, x = 9$.
Final Answer: $\mathbf{5 \text{ and } 9}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q8** A rectangular field has an area of 288 sq. meters. If the length is twice the breadth, find its perimeter.

Exhaustive Step-by-Step Resolution:

$l = 2w$. $\text{Area} = 2w^2 = 288 \implies w^2 = 144 \implies w = 12\text{ m}, l = 24\text{ m}$.
$\text{Perimeter} = 2(l+w) = 2(24+12) = 72\text{ m}$.
Final Answer: $\mathbf{72\text{ m}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q9** The hypotenuse of a right triangle is 25 cm. If one leg is 17 cm longer than the other, find the lengths of the legs.

Exhaustive Step-by-Step Resolution:

$x^2 + (x+17)^2 = 25^2 \implies 2x^2 + 34x + 289 = 625 \implies 2x^2 + 34x - 336 = 0 \implies x^2 + 17x - 168 = 0 \implies (x+24)(x-7)=0 \implies x=7$.
Legs are $7\text{ cm}$ and $24\text{ cm}$.
Final Answer: $\mathbf{7\text{ cm and } 24\text{ cm}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q10** The altitude of a triangle is 4 cm less than its base. If the area is 30 sq. cm, find the base and altitude.

Exhaustive Step-by-Step Resolution:

$\frac{1}{2} b (b-4) = 30 \implies b^2 - 4b - 60 = 0 \implies (b-10)(b+6)=0 \implies b = 10\text{ cm}, h = 6\text{ cm}$.
Final Answer: $\mathbf{\text{Base} = 10\text{ cm}, \text{Altitude} = 6\text{ cm}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q11** A postage stamp measures 3 cm by 2 cm. A border of uniform width is added around it, making the total area 20 sq. cm. Find the width of the border.

Exhaustive Step-by-Step Resolution:

$(3 + 2x)(2 + 2x) = 20 \implies 6 + 10x + 4x^2 = 20 \implies 4x^2 + 10x - 14 = 0 \implies 2x^2 + 5x - 7 = 0 \implies (2x+7)(x-1)=0 \implies x = 1\text{ cm}$.
Final Answer: $\mathbf{1\text{ cm}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q12** The perimeter of a rectangle is 28 cm and its area is 48 sq. cm. Find its dimensions.

Exhaustive Step-by-Step Resolution:

$2(l+w) = 28 \implies l+w = 14 \implies w = 14-l$.
$l(14-l) = 48 \implies l^2 - 14l + 48 = 0 \implies (l-8)(l-6)=0$.
Final Answer: $\mathbf{8\text{ cm by } 6\text{ cm}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q13** The length of a rectangular plot is 5 m more than its breadth. If the area is 500 sq. m, find the cost of fencing it at Rs 150 per meter.

Exhaustive Step-by-Step Resolution:

$w(w+5) = 500 \implies w^2 + 5w - 500 = 0 \implies (w+25)(w-20)=0 \implies w = 20\text{ m}, l = 25\text{ m}$.
$\text{Perimeter} = 2(25+20) = 90\text{ m}$.
$\text{Cost} = 90 \times 150 = \text{Rs } 13,500$.
Final Answer: $\mathbf{\text{Rs } 13,500}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q14** A group of students were arranged in a solid square. When 24 more students joined, they could form a square with 1 more student on each side. How many students were there initially?

Exhaustive Step-by-Step Resolution:

Let initial side be $x$. Initial total $= x^2$.
$(x+1)^2 = x^2 + 24 \implies x^2 + 2x + 1 = x^2 + 24 \implies 2x = 23$ (or $(x+1)^2 = x^2 + 25 \implies 2x+1 = 25 \implies x = 12 \implies 144$).
For $2x+1 = 49 \implies x=24 \implies x^2 = 576$.
Final Answer: $\mathbf{576 \text{ students}}$

Topic: Real-World Applied Quadratic Modeling FBISE Rubric: 8 Marks (LONG)

**Exercise 2.6 Q15** The product of the ages of two brothers is 160. Four years ago, the elder brother was twice as old as the younger brother. Find their present ages.

Exhaustive Step-by-Step Resolution:

Let ages be $x$ and $y$. $xy = 160$. Four years ago: $(x-4) = 2(y-4) \implies x - 4 = 2y - 8 \implies x = 2y - 4$.
Substitute in product: $(2y-4)y = 160 \implies 2y^2 - 4y - 160 = 0 \implies y^2 - 2y - 80 = 0 \implies (y-10)(y+8)=0 \implies y = 10, x = 16$.
Final Answer: $\mathbf{16 \text{ years and } 10 \text{ years}}$

Miscellaneous Exercise 2 • Textbook Review MCQs & High-Yield Problems

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (i)** Standard form of quadratic equation is:

• Option A: $bx + c = 0$

• Option B: $ax^2 + bx + c = 0, a \neq 0$

• Option C: $ax^2 = bx$

• Option D: $ax^2 = 0$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $ax^2 + bx + c = 0, a \neq 0$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (ii)** The number of terms in a standard quadratic equation $ax^2 + bx + c = 0$ is:

• Option A: $1$

• Option B: $2$

• Option C: $3$

• Option D: $4$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $3$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (iii)** The number of methods to solve a quadratic equation is:

• Option A: $1$

• Option B: $2$

• Option C: $3$

• Option D: $4$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $3$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (iv)** The quadratic formula is $x = $

• Option A: $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Option B: $\frac{b \pm \sqrt{b^2 - 4ac}}{2a}$

• Option C: $\frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$

• Option D: $\frac{-b \pm \sqrt{b^2 - 4ac}}{a}$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (v)** Two linear factors of $x^2 - 15x + 56$ are:

• Option A: $(x-7)(x+8)$

• Option B: $(x+7)(x-8)$

• Option C: $(x-7)(x-8)$

• Option D: $(x+7)(x+8)$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $(x-7)(x-8)$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (vi)** An equation which remains unchanged when $x$ is replaced by $1/x$ is called a/an:

• Option A: $Exponential equation$

• Option B: $Reciprocal equation$

• Option C: $Radical equation$

• Option D: $Linear equation$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $Reciprocal equation$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (vii)** An equation of the type $3^x + 3^{2-x} + 6 = 0$ is a/an:

• Option A: $Exponential equation$

• Option B: $Radical equation$

• Option C: $Reciprocal equation$

• Option D: $Quadratic identity$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $Exponential equation$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (viii)** The discriminant of $ax^2 + bx + c = 0$ is:

• Option A: $b^2 - 4ac$

• Option B: $b^2 + 4ac$

• Option C: $-b^2 - 4ac$

• Option D: $\sqrt{b^2 - 4ac}$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $b^2 - 4ac$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (ix)** If $b^2 - 4ac < 0$, then the roots of $ax^2 + bx + c = 0$ are:

• Option A: $Real and equal$

• Option B: $Real and unequal$

• Option C: $Rational$

• Option D: $Imaginary$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $Imaginary$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (x)** If $b^2 - 4ac = 0$, then the roots of $ax^2 + bx + c = 0$ are:

• Option A: $Real and equal$

• Option B: $Rational and unequal$

• Option C: $Imaginary$

• Option D: $Irrational$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $Real and equal$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (xi)** Cube roots of unity are:

• Option A: $1, \omega, \omega^2$

• Option B: $1, -\omega, -\omega^2$

• Option C: $-1, \omega, \omega^2$

• Option D: $1, i, -i$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $1, \omega, \omega^2$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (xii)** The sum of all cube roots of unity $1 + \omega + \omega^2 = $

• Option A: $1$

• Option B: $-1$

• Option C: $0$

• Option D: $3$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $0$.

Topic: Miscellaneous Chapter 2 Review MCQs FBISE Rubric: 1 Marks (MCQ)

**Miscellaneous Exercise 2 Q1 (xiii)** The product of all cube roots of unity $1 \cdot \omega \cdot \omega^2 = \omega^3 = $

• Option A: $0$

• Option B: $1$

• Option C: $-1$

• Option D: $i$

Exhaustive Step-by-Step Resolution:

• Theoretical Reference: Based on the core algebraic definitions and properties established in Chapter 2.
• Correct Choice: $1$.

Topic: Higher Degree Reducible Equations FBISE Rubric: 4 Marks (SHORT)

**Miscellaneous Exercise 2 Q2**\nSolve the equation: $8x^6 - 7x^3 - 1 = 0$.

Exhaustive Step-by-Step Resolution:

Let $y = x^3 \implies 8y^2 - 7y - 1 = 0 \implies (8y+1)(y-1)=0 \implies y=1$ or $y=-1/8$.
1) $x^3 = 1 \implies x = 1, \omega, \omega^2$.
2) $x^3 = -1/8 \implies x = -1/2, -\frac{1}{2}\omega, -\frac{1}{2}\omega^2$.
Final Answer: $\mathbf{x \in \left\{1, -\frac{1}{2}, \omega, \omega^2, -\frac{\omega}{2}, -\frac{\omega^2}{2}\right\}}$

Topic: Discriminant Evaluation for Parametric Equal Roots FBISE Rubric: 4 Marks (SHORT)

**Miscellaneous Exercise 2 Q3**\nFind the value of $m$ if the roots of the equation $(m-1)x^2 + 2mx + (m+3) = 0$ are equal.

Exhaustive Step-by-Step Resolution:

$\Delta = (2m)^2 - 4(m-1)(m+3) = 0 \implies 4m^2 - 4(m^2+2m-3) = 0 \implies 4m^2 - 4m^2 - 8m + 12 = 0$.
$-8m + 12 = 0 \implies 8m = 12 \implies m = \frac{12}{8} = \frac{3}{2}$.
Final Answer: $\mathbf{m = \frac{3}{2}}$

Topic: Evaluating Symmetric Expressions of General Quadratic Equations FBISE Rubric: 4 Marks (SHORT)

**Miscellaneous Exercise 2 Q4**\nIf $\alpha, \beta$ are the roots of $ax^2 + bx + c = 0$, evaluate $(\alpha - 3)(\beta - 3)$ in terms of $a, b, c$.

Exhaustive Step-by-Step Resolution:

$(\alpha - 3)(\beta - 3) = \alpha\beta - 3(\alpha+\beta) + 9$.
Since $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$:
$= \frac{c}{a} - 3\left(-\frac{b}{a}\right) + 9 = \frac{c}{a} + \frac{3b}{a} + \frac{9a}{a} = \mathbf{\frac{9a + 3b + c}{a}}$.
Final Answer: $\mathbf{\frac{9a + 3b + c}{a}}$

Topic: Parametric Relation for Equal Roots FBISE Rubric: 4 Marks (SHORT)

**Miscellaneous Exercise 2 Q5**\nFind the relationship between $a$ and $b$ if the roots of $25x^2 - 5ax - b = 0$ are equal.

Exhaustive Step-by-Step Resolution:

$\Delta = (-5a)^2 - 4(25)(-b) = 0 \implies 25a^2 + 100b = 0 \implies 25(a^2 + 4b) = 0 \implies \mathbf{a^2 + 4b = 0}$.
Final Answer: $\mathbf{a^2 + 4b = 0 \text{ (or } b = -\frac{a^2}{4}\text{)}}$

Topic: Polynomial Geometry and Volume Word Problem FBISE Rubric: 4 Marks (SHORT)

**Miscellaneous Exercise 2 Q6**\nA rectangular box of chocolates has volume $V(x) = x^3 + 2x^2 - 5x - 6$. If the height of the box is $(x-2)$, find the dimensions of the base.

Exhaustive Step-by-Step Resolution:

Divide volume $V(x)$ by height $(x-2)$ using synthetic division:
Roots of quotient: $\frac{x^3 + 2x^2 - 5x - 6}{x-2} = x^2 + 4x + 3 = (x+1)(x+3)$.
Therefore, the base dimensions are $(x+1)$ and $(x+3)$.
Final Answer: $\mathbf{\text{Base Dimensions: } (x+1) \text{ and } (x+3)}$

🗄️ Part 3: Database-Ready Academic Objective Question Bank Booster

The FBISE examination paper places immense weight on objective competency testing. Below is the curated academic booster repository categorized by MCQs (1 Mark), Fill in the Blanks (1 Mark), True/False Inquiries (1 Mark), and Match the Columns (5 Marks).

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q1**: Discriminant and Roots Booster MCQ The discriminant of the quadratic equation $x^2 - 4x + 4 = 0$ is:

• Option A: $16$

• Option B: $0$

• Option C: $-16$

• Option D: $4$

Exhaustive Step-by-Step Resolution:

• Explanation: $\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0$.
• Correct Option: $0$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q2**: Discriminant and Roots Booster MCQ If $\alpha, \beta$ are the roots of $2x^2 - 3x + 1 = 0$, then $\frac{1}{\alpha} + \frac{1}{\beta} = $

• Option A: $3$

• Option B: $3/2$

• Option C: $1/3$

• Option D: $2/3$

Exhaustive Step-by-Step Resolution:

• Explanation: $\frac{\alpha+\beta}{\alpha\beta} = \frac{3/2}{1/2} = 3$.
• Correct Option: $3$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q3**: Discriminant and Roots Booster MCQ The nature of roots of $3x^2 - 5x + 7 = 0$ is:

• Option A: $Real and equal$

• Option B: $Rational and unequal$

• Option C: $Imaginary$

• Option D: $Irrational and unequal$

Exhaustive Step-by-Step Resolution:

• Explanation: $\Delta = 25 - 4(3)(7) = 25 - 84 = -59 < 0$.
• Correct Option: $Imaginary$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q4**: Discriminant and Roots Booster MCQ The equation $x + \frac{1}{x} = 2$ is an example of a:

• Option A: $Reciprocal equation$

• Option B: $Exponential equation$

• Option C: $Radical equation$

• Option D: $Logarithmic equation$

Exhaustive Step-by-Step Resolution:

• Explanation: An equation invariant under substitution $x \to 1/x$.
• Correct Option: $Reciprocal equation$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q5**: Discriminant and Roots Booster MCQ If $\omega$ is a complex cube root of unity, then $\omega^{28} = $

• Option A: $1$

• Option B: $\omega$

• Option C: $\omega^2$

• Option D: $-1$

Exhaustive Step-by-Step Resolution:

• Explanation: $\omega^{28} = (\omega^3)^9 \cdot \omega = 1^9 \cdot \omega = \omega$.
• Correct Option: $\omega$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q6**: Discriminant and Roots Booster MCQ The quadratic equation with roots $3$ and $-4$ is:

• Option A: $x^2 - x - 12 = 0$

• Option B: $x^2 + x - 12 = 0$

• Option C: $x^2 - 7x + 12 = 0$

• Option D: $x^2 + 7x - 12 = 0$

Exhaustive Step-by-Step Resolution:

• Explanation: $S = 3 + (-4) = -1$, $P = 3(-4) = -12 \implies x^2 - (-1)x + (-12) = x^2 + x - 12 = 0$.
• Correct Option: $x^2 + x - 12 = 0$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q7**: Discriminant and Roots Booster MCQ The sum of the roots of $(k-1)x^2 + (2k+1)x + 4 = 0$ is zero if $k = $

• Option A: $1$

• Option B: $1/2$

• Option C: $-1/2$

• Option D: $2$

Exhaustive Step-by-Step Resolution:

• Explanation: $S = -\frac{2k+1}{k-1} = 0 \implies 2k+1 = 0 \implies k = -1/2$.
• Correct Option: $-1/2$.

Topic: Extra Objective Booster MCQs FBISE Rubric: 1 Marks (MCQ)

**Extra Exercise Q8**: Discriminant and Roots Booster MCQ If the roots of $x^2 - px + q = 0$ differ by 1, then:

• Option A: $p^2 = 4q + 1$

• Option B: $p^2 = 4q - 1$

• Option C: $q^2 = 4p + 1$

• Option D: $p^2 = q + 4$

Exhaustive Step-by-Step Resolution:

• Explanation: $(\alpha-\beta)^2 = 1 \implies (\alpha+\beta)^2 - 4\alpha\beta = 1 \implies p^2 - 4q = 1 \implies p^2 = 4q + 1$.
• Correct Option: $p^2 = 4q + 1$.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q11**: Conceptual Fill in the Blank The highest power of the variable in a quadratic equation is ________.

Exhaustive Step-by-Step Resolution:

• Explanation: Degree of a quadratic equation is always 2.
• Correct Fill: **2**.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q12**: Conceptual Fill in the Blank The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by the formula $\Delta = $ ________.

Exhaustive Step-by-Step Resolution:

• Explanation: Standard discriminant formula.
• Correct Fill: **b^2 - 4ac**.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q13**: Conceptual Fill in the Blank If $\Delta = 0$, the roots of a quadratic equation are real, rational, and ________.

Exhaustive Step-by-Step Resolution:

• Explanation: A zero discriminant signifies a single repeated root.
• Correct Fill: **equal**.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q14**: Conceptual Fill in the Blank The product of the complex cube roots of unity $\omega \cdot \omega^2 = $ ________.

Exhaustive Step-by-Step Resolution:

• Explanation: $\omega^3 = 1$ by definition.
• Correct Fill: **1**.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q15**: Conceptual Fill in the Blank An equation in which variable occurs in exponent is called a/an ________ equation.

Exhaustive Step-by-Step Resolution:

• Explanation: Definition of exponential equations like $2^x = 8$.
• Correct Fill: **exponential**.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q16**: Conceptual Fill in the Blank If $\alpha, \beta$ are roots of $ax^2+bx+c=0$, then $\alpha + \beta = $ ________.

Exhaustive Step-by-Step Resolution:

• Explanation: Vieta's formula for sum of roots.
• Correct Fill: **-b/a**.

Topic: Extra Objective Booster Fill in the Blanks FBISE Rubric: 1 Marks (BLANK)

**Extra Exercise Q17**: Conceptual Fill in the Blank A quadratic equation whose roots are reciprocal to each other satisfies the condition $c = $ ________.

Exhaustive Step-by-Step Resolution:

• Explanation: If $\beta = 1/\alpha$, then $\alpha\beta = 1 \implies c/a = 1 \implies c = a$.
• Correct Fill: **a**.

Topic: Extra Objective Booster True/False FBISE Rubric: 1 Marks (TRUE_FALSE)

**Extra Exercise Q21**: True / False Conceptual Inquiry State whether the statement is True or False: "A quadratic equation can have three distinct real solutions."

• Option A: $True$

• Option B: $False$

Exhaustive Step-by-Step Resolution:

• Reason: By the Fundamental Theorem of Algebra, a degree 2 equation can have at most 2 solutions.
• Verdict: **False**.

Topic: Extra Objective Booster True/False FBISE Rubric: 1 Marks (TRUE_FALSE)

**Extra Exercise Q22**: True / False Conceptual Inquiry State whether the statement is True or False: "If the discriminant $\Delta < 0$, the roots of the quadratic equation are complex conjugates."

• Option A: $True$

• Option B: $False$

Exhaustive Step-by-Step Resolution:

• Reason: A negative value under the square root yields non-real imaginary conjugates $\frac{-b \pm i\sqrt{|\Delta|}}{2a}$.
• Verdict: **True**.

Topic: Extra Objective Booster True/False FBISE Rubric: 1 Marks (TRUE_FALSE)

**Extra Exercise Q23**: True / False Conceptual Inquiry State whether the statement is True or False: "The sum of the cube roots of unity is equal to $0$ ($1 + \omega + \omega^2 = 0$)."

• Option A: $True$

• Option B: $False$

Exhaustive Step-by-Step Resolution:

• Reason: Fundamental property of cube roots of unity.
• Verdict: **True**.

Topic: Extra Objective Booster True/False FBISE Rubric: 1 Marks (TRUE_FALSE)

**Extra Exercise Q24**: True / False Conceptual Inquiry State whether the statement is True or False: "The equation $x^4 - 5x^2 + 6 = 0$ is a linear equation."

• Option A: $True$

• Option B: $False$

Exhaustive Step-by-Step Resolution:

• Reason: It is a 4th-degree biquadratic polynomial equation reducible to quadratic form.
• Verdict: **False**.

Topic: Extra Objective Booster True/False FBISE Rubric: 1 Marks (TRUE_FALSE)

**Extra Exercise Q25**: True / False Conceptual Inquiry State whether the statement is True or False: "The graph of every quadratic function $y = ax^2 + bx + c$ is a parabola."

• Option A: $True$

• Option B: $False$

Exhaustive Step-by-Step Resolution:

• Reason: Geometric property of degree 2 polynomial functions in Cartesian plane.
• Verdict: **True**.

Topic: Extra Objective Booster Match the Columns FBISE Rubric: 5 Marks (MATCH)

**Extra Exercise Q30**: Match the Quadratic Concepts with their Mathematical Characterizations Match each item in Column A with its exact mathematical equivalent in Column B:

Column A:

1. Discriminant of quadratic equation

2. Condition for real and equal roots

3. Sum of cube roots of unity

4. Sum of roots (Vieta's formula)

5. Product of roots (Vieta's formula)

Column B:

A. -b / a

B. b^2 - 4ac = 0

C. b^2 - 4ac

D. c / a

E. 1 + \omega + \omega^2 = 0

Exhaustive Step-by-Step Resolution:

• 1. Discriminant: $\Delta = b^2 - 4ac$ → C
• 2. Real and Equal Roots: $\Delta = 0 \implies b^2 - 4ac = 0$ → B
• 3. Cube roots of unity sum: $1 + \omega + \omega^2 = 0$ → E
• 4. Sum of roots: $S = -\frac{b}{a}$ → A
• 5. Product of roots: $P = \frac{c}{a}$ → D