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Quadratic Equations

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CAT 2025 Lesson : Quadratic Equations - Factorisation Method

There are three commonly used methods to solve quadratic equations. These are explained below.

This is used when the roots are small integers or fractions. Most of the questions will be of this type in CAT and other MBA entrance tests.

	Method	Example
Step 1	In the equation $ax^{2} + bx + c = 0$ , split b into $2$ parts such that sum of the parts = b and product of parts $=$ ac	In $6x^{2} + 7x + 2 = 0$ , $b = 7$ & $ac = 12$ $4 + 3 = 7$ and $4 \times 3 = 12$
Step 2	Factorise the equation by taking two terms at a time.	$6x^{2} + 7x + 2 = 0$ ⇒ $6x^{2} + 4x + 3x + 2 = 0$ ⇒ $2x (3x + 2) + 1 (3x + 2) = 0$ ⇒ $(2x + 1) (3x + 2) = 0$
Step 3	One of the factors has to equal 0 for the equation to be $0$ . Therefore, equate the factors to $0$ to find the $2$ roots.	⇒ $2x + 1 = 0$ or $3x + 2 = 0$ ⇒ $x = \dfrac{-1}{2}$ or $\dfrac{-2}{3}$

What are the roots of the equations

x^{2} - x - 6 = 0

and

12 x^{2} + 5x - 3 = 0

x^{2} - x - 6 = 0

(

b = -1

and

ac = -6

b

can be split as

-3

and

2

.)
⇒

x^{2} - 3x + 2x - 6 = 0

⇒

x(x - 3) + 2(x - 3) = 0

⇒

(x + 2) (x - 3) = 0

⇒

\bm{x = -2, 3}

12x^{2} + 5x - 3 = 0

(

b = 5

and

ac = -36

b

can be split as

9

and

-4

.)
⇒

12x^{2} + 9x - 4x - 3 = 0

⇒

3x (4x + 3) - 1(4x + 3) = 0

⇒

(3x - 1) (4x + 3) = 0

⇒

\bm{x = \dfrac{1}{3}, \dfrac{-3}{4}}

Alternatively (Recommended)

The following improvisation helps saves time and is, therefore, recommended.

Step 1: Split

b

into

2

terms as stated earlier (sum of terms

= b

and product of terms

= ac

)
Step 2: Change their signs.
Step 3: Divide them by

a

and these are your roots.

x^{2} - x - 6 = 0

(

b

can be split as

-3

and

2

.)
⇒

x = \dfrac{3}{1}, \dfrac{-2}{1} \bm{= 3, -2}

12x^{2} + 5x - 3 = 0

(

b

can be split as

9

and

-4

.)

⇒

x = \dfrac{-9}{12}, \dfrac{4}{12} = \bm{\dfrac{-3}{4}, \dfrac{1}{3}}

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