**2.4 Roots of Quadratic Equations**

**1.**A

**root of quadratic equation**is the value of the unknown which satisfies the quadratic equation.

**Roots of an equation are also called the**

2.

2.

**solution**of an equation.

**To solve a quadratic equation by the**

3.

3.

**factorisation**method, follow the steps below:

**Express the quadratic equation in general form**

*Step 1:**ax*

^{2}+

*bx*+

*c*= 0.

**Factorise the quadratic expression**

*Step 2:**ax*

^{2}+

*bx*+

*c*= 0 as the product of two linear expressions, that is, (

*mx*+

*p*) (

*nx*+

*q*) = 0.

**Equate each factor to zero and obtain the roots or solutions of the quadratic equation.**

*Step 3:***Example 1**:

Solve the quadratic equation
$\frac{2{x}^{2}-5}{3}=3x$

*Solution:*$\frac{2{x}^{2}-5}{3}=3x$

2

*x*^{2}– 5 = 9*x*2

*x*^{2}– 9*x –*5 = 0(

*x*– 5)(2*x*+ 1) = 0*x*– 5 = 0,

*x*= 5

or 2

$x=-\frac{1}{2}$
*x*+ 1 = 0Therefore,

*x*= 5 and*x*= ½are**roots or solutions**of the quadratic equation.**Example 2**:

Solve the quadratic equation 4

*x*^{2}– 12 = –13*x*

*Solution:*4

*x*^{2}– 12 = –13*x*4

*x*^{2}+ 13*x*– 12 = 0(4

*x*– 3)(*x*+ 4) = 04

*x*– 3 = 0, $x=\frac{3}{4}$or

*x*+ 4 = 0*x*= –4

**:**

Example 3

Example 3

Solve the quadratic equation 5

*x*^{2}= 3 (*x*+ 2) – 4

Solution:Solution:

5

*x*^{2}= 3 (*x*+ 2) – 45

*x*^{2}= 3*x*+ 6 – 45

*x*^{2}– 3*x*– 2 = 0(5

*x*+ 2)(*x*– 1) = 05

*x*+ 2 = 0, $x=-\frac{2}{5}$or

$\frac{3x(x-3)}{4}=-x+3$

*x*– 1 = 0*x*= 1

**Example 4**:

Solve the quadratic equation

$\frac{3x(x-3)}{4}=-x+3.$

*Solution:*$\frac{3x(x-3)}{4}=-x+3$

3

*x*^{2}– 9*x*= – 4*x*+ 123

*x*^{2}– 5*x*– 12 = 0(3

*x*+ 4)(*x*– 3) = 03

*x*+ 4 = 0, $x=-\frac{4}{3}$or

*x*– 3 = 0*x*= 3