Class 10 Mathematics - Ch 2: Quadratic Equations
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📘 Comprehensive Syllabus & Examination Guide
Class 10 Mathematics - Ch 2: Quadratic Equations
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Question Types & Curriculum Breakdown
Total Question Pool
100%
177 Questions
Combined Active Syllabus
Short
60%
107 Questions
Available
Long
20%
36 Questions
Available
Multiple Choice (MCQs)
12%
21 MCQs
Available
Blank
4%
7 Questions
Available
True / False
3%
5 Questions
Available
Match
1%
1 Questions
Available
📊 Question Pool Structure
177 Solved Questions (MCQs, Short & Long Questions, Blanks, True/False).
⚡ Recommended Pacing
1 to 3 minutes per question depending on question type (MCQ, Short, Long).
⚖️ Scoring & Negative Marking
1 to 5 marks per question aligned with official board examination rubrics.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Class 10 Mathematics - Ch 2: Quadratic Equations, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
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📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Standard Form of Quadratic Equations
MEDIUM • SHORT
**Exercise 2.1 Q1 (i)**
Write the following quadratic equation in standard form $ax^2 + bx + c = 0$:
$$(x+7)(x-3) = -7$$
Write the following quadratic equation in standard form $ax^2 + bx + c = 0$:
$$(x+7)(x-3) = -7$$
✓ Correct Answer: x^2 + 4x - 14 = 0
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Expand the product):</strong> $(x+7)(x-3) = x^2 - 3x + 7x - 21 = x^2 + 4x - 21$.<br/>
• <strong>Step 2 (Shift constant to left):</strong> $x^2 + 4x - 21 + 7 = 0 \implies \mathbf{x^2 + 4x - 14 = 0}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{x^2 + 4x - 14 = 0}$ (Standard quadratic form where $a=1, b=4, c=-14$).
• <strong>Step 2 (Shift constant to left):</strong> $x^2 + 4x - 21 + 7 = 0 \implies \mathbf{x^2 + 4x - 14 = 0}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{x^2 + 4x - 14 = 0}$ (Standard quadratic form where $a=1, b=4, c=-14$).
Sample Question 2
Standard Form of Quadratic Equations
MEDIUM • SHORT
**Exercise 2.1 Q1 (ii)**
Write the following quadratic equation in standard form $ax^2 + bx + c = 0$:
$$\frac{x^2+4}{3} - \frac{x}{7} = 1$$
Write the following quadratic equation in standard form $ax^2 + bx + c = 0$:
$$\frac{x^2+4}{3} - \frac{x}{7} = 1$$
✓ Correct Answer: 7x^2 - 3x + 7 = 0
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Clear denominators by multiplying by $\text{LCM}(3,7)=21$):</strong> $21 \left(\frac{x^2+4}{3}\right) - 21 \left(\frac{x}{7}\right) = 21(1)$.<br/>
• <strong>Step 2 (Simplify terms):</strong> $7(x^2+4) - 3(x) = 21 \implies 7x^2 + 28 - 3x = 21$.<br/>
• <strong>Step 3 (Rearrange into standard form):</strong> $7x^2 - 3x + 28 - 21 = 0 \implies \mathbf{7x^2 - 3x + 7 = 0}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{7x^2 - 3x + 7 = 0}$
• <strong>Step 2 (Simplify terms):</strong> $7(x^2+4) - 3(x) = 21 \implies 7x^2 + 28 - 3x = 21$.<br/>
• <strong>Step 3 (Rearrange into standard form):</strong> $7x^2 - 3x + 28 - 21 = 0 \implies \mathbf{7x^2 - 3x + 7 = 0}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{7x^2 - 3x + 7 = 0}$
Sample Question 3
Standard Form of Quadratic Equations
MEDIUM • SHORT
**Exercise 2.1 Q1 (iii)**
Write the following quadratic equation in standard form $ax^2 + bx + c = 0$:
$$\frac{x}{x+1} + \frac{x+1}{x} = 6$$
Write the following quadratic equation in standard form $ax^2 + bx + c = 0$:
$$\frac{x}{x+1} + \frac{x+1}{x} = 6$$
✓ Correct Answer: 4x^2 + 4x - 1 = 0
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Multiply the entire equation by $x(x+1)$):</strong> $x(x) + (x+1)^2 = 6x(x+1)$.<br/>
• <strong>Step 2 (Expand all terms):</strong> $x^2 + (x^2 + 2x + 1) = 6x^2 + 6x \implies 2x^2 + 2x + 1 = 6x^2 + 6x$.<br/>
• <strong>Step 3 (Collect all terms on one side):</strong> $6x^2 - 2x^2 + 6x - 2x - 1 = 0 \implies \mathbf{4x^2 + 4x - 1 = 0}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{4x^2 + 4x - 1 = 0}$
• <strong>Step 2 (Expand all terms):</strong> $x^2 + (x^2 + 2x + 1) = 6x^2 + 6x \implies 2x^2 + 2x + 1 = 6x^2 + 6x$.<br/>
• <strong>Step 3 (Collect all terms on one side):</strong> $6x^2 - 2x^2 + 6x - 2x - 1 = 0 \implies \mathbf{4x^2 + 4x - 1 = 0}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{4x^2 + 4x - 1 = 0}$
Sample Question 4
Solving Quadratic Equations by Factoring
MEDIUM • SHORT
**Exercise 2.1 Q2 (i)**
Solve the following quadratic equation by factorization:
$$x^2 - x - 20 = 0$$
Solve the following quadratic equation by factorization:
$$x^2 - x - 20 = 0$$
✓ Correct Answer: x \in \{-4, 5\}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Find factors of product $-20$ with sum $-1$):</strong> Factors are $-5$ and $+4$.<br/>
• <strong>Step 2 (Split middle term and factor by grouping):</strong> $x^2 - 5x + 4x - 20 = 0 \implies x(x-5) + 4(x-5) = 0 \implies (x-5)(x+4) = 0$.<br/>
• <strong>Step 3 (Apply Zero Product Property):</strong> $x - 5 = 0 \implies x = 5$ or $x + 4 = 0 \implies x = -4$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\{-4, 5\}}$
• <strong>Step 2 (Split middle term and factor by grouping):</strong> $x^2 - 5x + 4x - 20 = 0 \implies x(x-5) + 4(x-5) = 0 \implies (x-5)(x+4) = 0$.<br/>
• <strong>Step 3 (Apply Zero Product Property):</strong> $x - 5 = 0 \implies x = 5$ or $x + 4 = 0 \implies x = -4$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\{-4, 5\}}$
Sample Question 5
Solving Quadratic Equations by Factoring
MEDIUM • SHORT
**Exercise 2.1 Q2 (ii)**
Solve the following quadratic equation by factorization:
$$3y^2 = y(y-5)$$
Solve the following quadratic equation by factorization:
$$3y^2 = y(y-5)$$
✓ Correct Answer: y \in \left\{0, -\frac{5}{2}\right\}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Expand and bring all terms to LHS):</strong> $3y^2 = y^2 - 5y \implies 3y^2 - y^2 + 5y = 0 \implies 2y^2 + 5y = 0$.<br/>
• <strong>Step 2 (Factor out common term $y$):</strong> $y(2y + 5) = 0$.<br/>
• <strong>Step 3 (Solve for $y$):</strong> $y = 0$ or $2y + 5 = 0 \implies y = -\frac{5}{2}$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\left\{0, -\frac{5}{2}\right\}}$
• <strong>Step 2 (Factor out common term $y$):</strong> $y(2y + 5) = 0$.<br/>
• <strong>Step 3 (Solve for $y$):</strong> $y = 0$ or $2y + 5 = 0 \implies y = -\frac{5}{2}$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\left\{0, -\frac{5}{2}\right\}}$
Sample Question 6
Solving Quadratic Equations by Factoring
MEDIUM • SHORT
**Exercise 2.1 Q2 (iii)**
Solve the following quadratic equation by factorization:
$$4 - 32x = 17x^2$$
Solve the following quadratic equation by factorization:
$$4 - 32x = 17x^2$$
✓ Correct Answer: x \in \left\{-2, \frac{2}{17}\right\}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Rearrange into standard form):</strong> $17x^2 + 32x - 4 = 0$.<br/>
• <strong>Step 2 (Find product $17 \times (-4) = -68$ with sum $+32$):</strong> Factors are $+34$ and $-2$.<br/>
• <strong>Step 3 (Factor by grouping):</strong> $17x^2 + 34x - 2x - 4 = 0 \implies 17x(x+2) - 2(x+2) = 0 \implies (17x - 2)(x + 2) = 0$.<br/>
• <strong>Step 4 (Solve for $x$):</strong> $x + 2 = 0 \implies x = -2$ or $17x - 2 = 0 \implies x = \frac{2}{17}$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\left\{-2, \frac{2}{17}\right\}}$
• <strong>Step 2 (Find product $17 \times (-4) = -68$ with sum $+32$):</strong> Factors are $+34$ and $-2$.<br/>
• <strong>Step 3 (Factor by grouping):</strong> $17x^2 + 34x - 2x - 4 = 0 \implies 17x(x+2) - 2(x+2) = 0 \implies (17x - 2)(x + 2) = 0$.<br/>
• <strong>Step 4 (Solve for $x$):</strong> $x + 2 = 0 \implies x = -2$ or $17x - 2 = 0 \implies x = \frac{2}{17}$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\left\{-2, \frac{2}{17}\right\}}$
Sample Question 7
Solving Quadratic Equations by Factoring
MEDIUM • SHORT
**Exercise 2.1 Q2 (iv)**
Solve the following quadratic equation by factorization:
$$x^2 - 11x = 152$$
Solve the following quadratic equation by factorization:
$$x^2 - 11x = 152$$
✓ Correct Answer: x \in \{-8, 19\}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Standard form):</strong> $x^2 - 11x - 152 = 0$.<br/>
• <strong>Step 2 (Find product $-152$ with sum $-11$):</strong> Factors are $-19$ and $+8$ (since $-19 \times 8 = -152$).<br/>
• <strong>Step 3 (Factor):</strong> $(x - 19)(x + 8) = 0$.<br/>
• <strong>Step 4 (Solve):</strong> $x = 19$ or $x = -8$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\{-8, 19\}}$
• <strong>Step 2 (Find product $-152$ with sum $-11$):</strong> Factors are $-19$ and $+8$ (since $-19 \times 8 = -152$).<br/>
• <strong>Step 3 (Factor):</strong> $(x - 19)(x + 8) = 0$.<br/>
• <strong>Step 4 (Solve):</strong> $x = 19$ or $x = -8$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\{-8, 19\}}$
Sample Question 8
Solving Quadratic Equations by Factoring
MEDIUM • SHORT
**Exercise 2.1 Q2 (v)**
Solve the following quadratic equation by factorization:
$$\frac{x+1}{x} + \frac{x}{x+1} = \frac{25}{12}$$
Solve the following quadratic equation by factorization:
$$\frac{x+1}{x} + \frac{x}{x+1} = \frac{25}{12}$$
✓ Correct Answer: x \in \left\{3, -\frac{4}{3}\right\} \text{ or } \left\{-\frac{4}{7}, \dots\right\}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Clear denominators by multiplying by $12x(x+1)$):</strong> $12(x+1)^2 + 12x^2 = 25x(x+1)$.<br/>
• <strong>Step 2 (Expand and simplify):</strong> $12(x^2 + 2x + 1) + 12x^2 = 25x^2 + 25x \implies 24x^2 + 24x + 12 = 25x^2 + 25x$.<br/>
• <strong>Step 3 (Rearrange):</strong> $25x^2 - 24x^2 + 25x - 24x - 12 = 0 \implies x^2 + x - 12 = 0$.<br/>
• <strong>Step 4 (Factorize):</strong> $(x + 4)(x - 3) = 0 \implies x = 3$ or $x = -4$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\{-4, 3\}}$
• <strong>Step 2 (Expand and simplify):</strong> $12(x^2 + 2x + 1) + 12x^2 = 25x^2 + 25x \implies 24x^2 + 24x + 12 = 25x^2 + 25x$.<br/>
• <strong>Step 3 (Rearrange):</strong> $25x^2 - 24x^2 + 25x - 24x - 12 = 0 \implies x^2 + x - 12 = 0$.<br/>
• <strong>Step 4 (Factorize):</strong> $(x + 4)(x - 3) = 0 \implies x = 3$ or $x = -4$.<br/>
• <strong>Final Answer:</strong> $\text{Solution Set} = \mathbf{\{-4, 3\}}$
Sample Question 9
Completing the Square Method
MEDIUM • SHORT
**Exercise 2.1 Q3 (i)**
Solve the following equation by completing the square method:
$$7x^2 + 2x - 1 = 0$$
Solve the following equation by completing the square method:
$$7x^2 + 2x - 1 = 0$$
✓ Correct Answer: x = \frac{-1 \pm 2\sqrt{2}}{7}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Divide by leading coefficient $7$):</strong> $x^2 + \frac{2}{7}x - \frac{1}{7} = 0 \implies x^2 + \frac{2}{7}x = \frac{1}{7}$.<br/>
• <strong>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):</strong> $x^2 + \frac{2}{7}x + \frac{1}{49} = \frac{1}{7} + \frac{1}{49}$.<br/>
• <strong>Step 3 (Complete the square on LHS):</strong> $\left(x + \frac{1}{7}\right)^2 = \frac{7+1}{49} = \frac{8}{49}$.<br/>
• <strong>Step 4 (Take square root on both sides):</strong> $x + \frac{1}{7} = \pm \frac{\sqrt{8}}{7} = \pm \frac{2\sqrt{2}}{7} \implies x = \frac{-1 \pm 2\sqrt{2}}{7}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{x = \frac{-1 \pm 2\sqrt{2}}{7}}$
• <strong>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):</strong> $x^2 + \frac{2}{7}x + \frac{1}{49} = \frac{1}{7} + \frac{1}{49}$.<br/>
• <strong>Step 3 (Complete the square on LHS):</strong> $\left(x + \frac{1}{7}\right)^2 = \frac{7+1}{49} = \frac{8}{49}$.<br/>
• <strong>Step 4 (Take square root on both sides):</strong> $x + \frac{1}{7} = \pm \frac{\sqrt{8}}{7} = \pm \frac{2\sqrt{2}}{7} \implies x = \frac{-1 \pm 2\sqrt{2}}{7}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{x = \frac{-1 \pm 2\sqrt{2}}{7}}$
Sample Question 10
Completing the Square Method
MEDIUM • SHORT
**Exercise 2.1 Q3 (ii)**
Solve the following equation by completing the square method:
$$ax^2 + 4x - a = 0 \quad (a \neq 0)$$
Solve the following equation by completing the square method:
$$ax^2 + 4x - a = 0 \quad (a \neq 0)$$
✓ Correct Answer: x = \frac{-2 \pm \sqrt{4+a^2}}{a}
📖 Step-by-Step Solution & Conceptual Rationale:
• <strong>Step 1 (Divide by $a$):</strong> $x^2 + \frac{4}{a}x - 1 = 0 \implies x^2 + \frac{4}{a}x = 1$.<br/>
• <strong>Step 2 (Add $\left(\frac{2}{a}\right)^2 = \frac{4}{a^2}$ to both sides):</strong> $x^2 + \frac{4}{a}x + \frac{4}{a^2} = 1 + \frac{4}{a^2} = \frac{a^2+4}{a^2}$.<br/>
• <strong>Step 3 (Complete the square):</strong> $\left(x + \frac{2}{a}\right)^2 = \frac{a^2+4}{a^2}$.<br/>
• <strong>Step 4 (Take square root):</strong> $x + \frac{2}{a} = \pm \frac{\sqrt{a^2+4}}{a} \implies x = \frac{-2 \pm \sqrt{a^2+4}}{a}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{x = \frac{-2 \pm \sqrt{a^2+4}}{a}}$
• <strong>Step 2 (Add $\left(\frac{2}{a}\right)^2 = \frac{4}{a^2}$ to both sides):</strong> $x^2 + \frac{4}{a}x + \frac{4}{a^2} = 1 + \frac{4}{a^2} = \frac{a^2+4}{a^2}$.<br/>
• <strong>Step 3 (Complete the square):</strong> $\left(x + \frac{2}{a}\right)^2 = \frac{a^2+4}{a^2}$.<br/>
• <strong>Step 4 (Take square root):</strong> $x + \frac{2}{a} = \pm \frac{\sqrt{a^2+4}}{a} \implies x = \frac{-2 \pm \sqrt{a^2+4}}{a}$.<br/>
• <strong>Final Answer:</strong> $\mathbf{x = \frac{-2 \pm \sqrt{a^2+4}}{a}}$
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