Class 10 Maths Chapter 4 – Quadratic Equations | NCERT Solutions, Methods & CBSE Board Questions (2025–26)

 Class 10 CBSE Maths – Chapter 4: Quadratic Equations

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🔹 Key Concepts and Formulas

✅ What is a Quadratic Equation?

A quadratic equation in the variable $x$ is of the form:
$ax^2 + bx + c = 0$
where:

• $a, b, c$ are real numbers

• $a ≠ 0$

It is called a quadratic equation because the highest exponent of the variable $x$ is 2.

✅ Standard Forms

• Pure quadratic: $ax^2 + c = 0$

• Quadratic trinomials: $ax^2 + bx + c = 0$

• Factorable quadratics: can be written as product of binomials.

✅ Methods of Solving Quadratic Equations

1. Factorisation Method

   • Express as a product of two linear factors

   • Set each factor = 0 and solve

2. Completing the Square Method

   • Convert to perfect square by adding/subtracting a constant

   • Take square root on both sides

3. Quadratic Formula:
   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
   This works for all quadratic equations.

✅ Discriminant (D) and Nature of Roots

$D = b^2 - 4ac$

• $D > 0$: Two real and distinct roots

• $D = 0$: Two real and equal roots

• $D < 0$: No real roots (complex roots)

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📘 NCERT Solved Examples (Key Highlights)

Example 1:

Solve: $x^2 + 5x + 6 = 0$
$x^2 + 2x + 3x + 6 = 0 \Rightarrow (x + 2)(x + 3) = 0 \Rightarrow x = -2, -3$

Example 2:

Solve using quadratic formula: $2x^2 + 3x - 2 = 0$
$D = 9 + 16 = 25 \Rightarrow x = \frac{-3 \pm 5}{4} = -2, \frac{1}{2}$

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📄 Exercise 4.1 – Check for Quadratic Equations

Q1. Check whether the following are quadratic equations:

(i) $x^2 - 3x + 1 = 0$ ✅ Yes
(ii) $x + \frac{1}{x} = 2$ ⛔ No
(iii) $(x - 2)^2 = 4$ ✅ Yes
(iv) $x^3 - x = 0$ ⛔ No (Degree is 3)

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📄 Exercise 4.2 – Solving by Factorisation

Q1. Solve:

(i) $x^2 + 7x + 10 = 0$ → $(x + 2)(x + 5) = 0 \Rightarrow x = -2, -5$
(ii) $2x^2 + 5x - 3 = 0$ → Split middle term: $(2x - 1)(x + 3) = 0 \Rightarrow x = \frac{1}{2}, -3$
(iii) $x^2 - 3x - 10 = 0$ → $(x - 5)(x + 2) = 0 \Rightarrow x = 5, -2$

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📄 Exercise 4.3 – Completing the Square

Q1. Solve: $x^2 + 6x - 7 = 0$

Add and subtract 9:
$x^2 + 6x + 9 - 9 - 7 = 0 \Rightarrow (x + 3)^2 = 16 \Rightarrow x + 3 = \pm 4 \Rightarrow x = 1, -7$

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📄 Exercise 4.4 – Quadratic Formula

Q1. Solve: $x^2 - 2x - 8 = 0$

$D = 4 + 32 = 36 \Rightarrow x = \frac{2 \pm 6}{2} = 4, -2$

Q2. $2x^2 - 4x + 3 = 0$ → $D = -8$ ⛔ No real roots

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📄 Exercise 4.5 – Nature of Roots

Q1. Determine the nature of roots:

(i) $x^2 + 4x + 4 = 0$ → $D = 0$: Real and equal
(ii) $x^2 + 2x + 5 = 0$ → $D = 4 - 20 = -16$: No real roots
(iii) $x^2 - 7x + 10 = 0$ → $D = 49 - 40 = 9$: Real and distinct

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🔁 Summary and Revision Notes

• Quadratic equations have degree 2

• Use factorisation when roots are rational and easy

• Use completing square or formula for complex cases

• Always check discriminant to understand root type

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📘 CBSE Previous Year Board Questions – Quadratic Equations

✅ 1-Mark

(2020) What is the value of discriminant for the equation $x^2 + 4x + 4 = 0$?
📌 Answer: D = $4^2 - 4×1×4 = 0$

(2019) Write nature of roots of $x^2 - 4x + 5 = 0$.
📌 D = $(-4)^2 - 4×1×5 = 16 - 20 = -4$ → No real roots

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✅ 2-Marks

(2020) Solve using quadratic formula: $2x^2 - 7x + 3 = 0$
📌 D = 49 - 24 = 25
$x = \frac{7 \pm 5}{4} \Rightarrow x = 3, \frac{1}{2}$

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✅ 3-Marks

(2018) Find roots of: $x^2 + x - 6 = 0$ by factorisation
📌 $(x + 3)(x - 2) = 0 \Rightarrow x = -3, 2$

(2016) Determine whether real roots exist for $x^2 + 2x + 5 = 0$
📌 D = 4 - 20 = -16 → No real roots

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✅ 4-Marks Word Problem

(2019) The product of two consecutive positive integers is 132. Find the integers.
📌 Let integers = x, x+1 → $x(x+1) = 132 \Rightarrow x^2 + x - 132 = 0$
Solve to get: $x = 11, x+1 = 12$

(2021 Sample) The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
📌 Let numbers be $x$ and $x + 2$:
$x^2+(x+2{)}^2=394\Rightarrow 2x^2+4x+4=394\Rightarrow x^2+2x-195=0\Rightarrow x=13\Rightarrow Numbers:13,15$