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Quadratic Equations

Quadratic Equations

MODULES

Basics of Polynomial & Quadratic Equations
Discriminant & Graphical Representation
Sum & Product of Roots
Factorisation Method
Formulation & Completion of Squares
Changes to Roots
Mistakes in Roots, Common Roots & Squaring
Infinite Series & Transposed
Other Types
Higher Order Equations
Synthetic Division & Remainder Theorem
Maxima, Minima & Descrates Rule
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Quadratic Equations 1
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Quadratic Equations 2
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Quadratic Equations 3
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PRACTICE

Quadratic Equations : Level 1
Quadratic Equations : level 2
Quadratic Equations : level 3
ALL MODULES

CAT 2025 Lesson : Quadratic Equations - Factorisation Method

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3.4 Solving Equations

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

3.4.1 Factorisation Method

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 ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0, split b into 222 parts such that sum of the parts = b and product of parts === ac In 6x2+7x+2=06x^{2} + 7x + 2 = 06x2+7x+2=0, b=7b = 7b=7 & ac=12ac = 12ac=12
4+3=74 + 3 = 74+3=7 and 4×3=124 \times 3 = 124×3=12
Step 2 Factorise the equation by taking two terms at a time. 6x2+7x+2=06x^{2} + 7x + 2 = 06x2+7x+2=0
⇒ 6x2+4x+3x+2=0 6x^{2} + 4x + 3x + 2 = 06x2+4x+3x+2=0
⇒ 2x(3x+2)+1(3x+2)=0 2x (3x + 2) + 1 (3x + 2) = 02x(3x+2)+1(3x+2)=0
⇒ (2x+1)(3x+2)=0 (2x + 1) (3x + 2) = 0(2x+1)(3x+2)=0
Step 3 One of the factors has to equal 0 for the equation to be 000. Therefore, equate the factors to 000 to find the 222 roots. ⇒ 2x+1=0 2x + 1 = 02x+1=0 or 3x+2=03x + 2 = 03x+2=0
⇒ x=−12 x = \dfrac{-1}{2}x=2−1​ or −23\dfrac{-2}{3}3−2​


Example 6

What are the roots of the equations x2−x−6=0x^{2} - x - 6 = 0x2−x−6=0 and 12x2+5x−3=012 x^{2} + 5x - 3 = 012x2+5x−3=0?

Solution

x2−x−6=0x^{2} - x - 6 = 0x2−x−6=0 (b=−1b = -1b=−1 and ac=−6ac = -6ac=−6. bbb can be split as −3-3−3 and 222.)
⇒ x2−3x+2x−6=0 x^{2} - 3x + 2x - 6 = 0x2−3x+2x−6=0
⇒ x(x−3)+2(x−3)=0x(x - 3) + 2(x - 3) = 0x(x−3)+2(x−3)=0
⇒ (x+2)(x−3)=0 (x + 2) (x - 3) = 0(x+2)(x−3)=0
⇒ x=−2,3\bm{x = -2, 3}x=−2,3

12x2+5x−3=012x^{2} + 5x - 3 = 012x2+5x−3=0 (b=5b = 5b=5 and ac=−36ac = -36ac=−36. bbb can be split as 999 and −4-4−4.)
⇒ 12x2+9x−4x−3=0 12x^{2} + 9x - 4x - 3 = 012x2+9x−4x−3=0
⇒ 3x(4x+3)−1(4x+3)=0 3x (4x + 3) - 1(4x + 3) = 03x(4x+3)−1(4x+3)=0
⇒ (3x−1)(4x+3)=0 (3x - 1) (4x + 3) = 0(3x−1)(4x+3)=0
⇒ x=13,−34\bm{x = \dfrac{1}{3}, \dfrac{-3}{4}}x=31​,4−3​

Alternatively (Recommended)

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

Step 1: Split bbb into 222 terms as stated earlier (sum of terms =b= b=b and product of terms =ac= ac=ac)
Step 2: Change their signs.
Step 3: Divide them by aaa and these are your roots.

x2−x−6=0x^{2} - x - 6 = 0x2−x−6=0 (bbb can be split as −3-3−3 and 222.)
⇒ x=31,−21=3,−2 x = \dfrac{3}{1}, \dfrac{-2}{1} \bm{= 3, -2}x=13​,1−2​=3,−2

12x2+5x−3=012x^{2} + 5x - 3 = 012x2+5x−3=0 (bbb can be split as 999 and −4-4−4.)

⇒ x=−912,412=−34,13 x = \dfrac{-9}{12}, \dfrac{4}{12} = \bm{\dfrac{-3}{4}, \dfrac{1}{3}}x=12−9​,124​=4−3​,31​


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