Solution of Quadratic Equations by Factorisation for CBSE Class 10

Once a real-world problem is formulated into a quadratic equation $ax^2 + bx + c = 0$, the next step is finding the numerical values of the variable that satisfy it—its roots. The earliest and most fundamental algebraic method for solving quadratic equations is the method of factorisation (often called splitting the middle term).

In CBSE Class 10 Mathematics, solving quadratic equations by factorisation is a core technique tested in both short 2-mark questions and multi-step 3-mark questions. While factoring simple integers is familiar from Class 9, Class 10 board exams frequently feature quadratic equations with irrational (radical) coefficients and literal algebraic constants.

What You Will Learn

• The Zero Product Property and how it yields roots
• Step-by-step procedure for splitting the middle term
• The product-sum rule: finding factors $p$ and $q$ such that $p + q = b$ and $p \times q = ac$
• Solving quadratic equations containing square roots ($\sqrt{2}, \sqrt{3}, \sqrt{5}$)
• Factoring equations with literal constants (e.g., $a^2 - b^2$)
• High-yield CBSE board examination questions and error-prevention tips

1. The Zero Product Property

The entire logical foundation of solving equations by factorisation rests on a simple property of arithmetic:

Zero Product Rule

If the product of two real numbers (or algebraic expressions) is zero, then at least one of the numbers must be equal to zero.

$\text{If } A \times B = 0, \quad \text{then either } A = 0 \quad \text{or} \quad B = 0 \quad \text{(or both)}.$

Therefore, if we can factor a quadratic equation into two linear factors: $(px + q)(rx + s) = 0$ then either: $px + q = 0 \implies x = -\frac{q}{p} \quad \text{or} \quad rx + s = 0 \implies x = -\frac{s}{r}$

2. Step-by-Step Method: Splitting the Middle Term

To solve $ax^2 + bx + c = 0$:

1. Step 1 (Identify $a, b, c$): Ensure the equation is written in standard form $ax^2 + bx + c = 0$.
2. Step 2 (Find Product $ac$): Calculate the product of the leading coefficient $a$ and the constant term $c$, i.e., $P = a \times c$.
3. Step 3 (Find Split Factors): Search for two numbers $p$ and $q$ such that:
   • Their sum equals the middle coefficient: $p + q = b$
   • Their product equals $ac$: $p \times q = a \times c$
4. Step 4 (Split the Middle Term): Rewrite the middle term $bx$ as $px + qx$: $ax^2 + px + qx + c = 0$
5. Step 5 (Group in Pairs): Group the first two terms and the last two terms to extract common factors: $x(ax + p) + \dots = 0$
6. Step 6 (Apply Zero Product Rule): Set each linear factor to zero to obtain the two roots.

3. Solved Step-by-Step Examples

Solved Example 1: Standard Integer Equation

Problem: Find the roots of the quadratic equation $6x^2 - x - 2 = 0$.

Solution:

1. Here $a = 6, b = -1, c = -2$.
2. Compute $ac = 6 \times (-2) = -12$.
3. We need two numbers $p$ and $q$ such that $p \times q = -12$ and $p + q = -1$. The factors of $-12$ that add to $-1$ are $-4$ and $+3$.
4. Split the middle term $-x$ as $-4x + 3x$: $6x^2 - 4x + 3x - 2 = 0$
5. Group by pairs: $2x(3x - 2) + 1(3x - 2) = 0$
6. Factor out the common binomial $(3x - 2)$: $(3x - 2)(2x + 1) = 0$
7. Apply the zero product property: $3x - 2 = 0 \implies x = \frac{2}{3}$ $2x + 1 = 0 \implies x = -\frac{1}{2}$
8. Therefore, <u>the roots are $x = \frac{2}{3}$ and $x = -\frac{1}{2}$</u>.

Solved Example 2: Radical (Square Root) Coefficients (CBSE Board Classic)

Problem: Find the roots of the quadratic equation $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.

Solution:

1. Identify coefficients: $a = \sqrt{2}, b = 7, c = 5\sqrt{2}$.
2. Compute product $ac$: $ac = (\sqrt{2}) \times (5\sqrt{2}) = 5 \times 2 = 10$
3. We need two numbers whose product is $10$ and whose sum is $7$. The numbers are $5$ and $2$ ($5 \times 2 = 10$ and $5 + 2 = 7$).
4. Split the middle term $7x$ as $5x + 2x$: $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$
5. Notice that $2 = \sqrt{2} \times \sqrt{2}$. Group terms carefully: $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$
6. Factor out the common binomial $(\sqrt{2}x + 5)$: $(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$
7. Set each factor to zero: $\sqrt{2}x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$ $x + \sqrt{2} = 0 \implies x = -\sqrt{2}$
8. Thus, <u>the roots are $x = -\sqrt{2}$ and $x = -\frac{5\sqrt{2}}{2}$</u>.

Solved Example 3: Repeated Middle Root with Radicals

Problem: Solve for $x$: $3x^2 - 2\sqrt{6}x + 2 = 0$.

Solution:

1. Here $a = 3, b = -2\sqrt{6}, c = 2$.
2. Product $ac = 3 \times 2 = 6$.
3. We need two numbers whose product is $6$ and sum is $-2\sqrt{6}$. The numbers are $-\sqrt{6}$ and $-\sqrt{6}$: $(-\sqrt{6}) \times (-\sqrt{6}) = 6, \quad (-\sqrt{6}) + (-\sqrt{6}) = -2\sqrt{6}$
4. Split the middle term: $3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0$
5. Recognize that $3 = \sqrt{3} \times \sqrt{3}$, $\sqrt{6} = \sqrt{3} \times \sqrt{2}$, and $2 = \sqrt{2} \times \sqrt{2}$: $\sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0$
6. Factor out $(\sqrt{3}x - \sqrt{2})$: $(\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0$
7. Setting both factors to zero gives: $x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}, \quad x = \sqrt{\frac{2}{3}}$
8. Therefore, the equation has two equal real roots: <u>$x = \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$</u>.

4. Summary and Examination Tips

Exam Tip: When factoring equations with radical terms like $\sqrt{2}x$ or $\sqrt{3}x$, always express whole numbers as products of square roots (e.g., $2 = \sqrt{2} \cdot \sqrt{2}$, $3 = \sqrt{3} \cdot \sqrt{3}$) to make the common binomial factor obvious.

Common Mistake: Forgetting to state both roots when they are identical! If $(x - 3)^2 = 0$, do not write just "$x = 3$". Write "$x = 3, 3$ (two equal roots)", as a quadratic equation must always have two roots.