 # 1.4 Roots of Quadratic Equations

1.4 Roots of Quadratic Equations
1. A root of quadratic equation is the value of the unknown which satisfies the quadratic equation.

2.
Roots of an equation are also called the solution of an equation.

3.
To solve a quadratic equation by the factorisation method, follow the steps below:

Step 1: Express the quadratic equation in general form ax2 + bx + c = 0.
Step 2: Factorise the quadratic expression ax2 + bx + c = 0 as the product of two linear expressions, that is, (mx+ p) (nx + q) = 0.
Step 3: Equate each factor to zero and obtain the roots or solutions of the quadratic equation.

Example 1:
Determine whether 1, 2, and 3 are the roots of the quadratic equation ${x}^{2}-5x+6=0$ .

Solution:
When x = 1,
$\begin{array}{l}{x}^{2}-5x+6=0\\ \left(1{\right)}^{2}-5\left(1\right)+6=0\\ 2=0\end{array}$
x = 1 does not satisfy the equation

When x = 2,
$\begin{array}{l}{x}^{2}-5x+6=0\\ \left(2{\right)}^{2}-5\left(2\right)+6=0\\ 0=0\end{array}$
x = 2 satisfies the equation.

When x = 3
$\begin{array}{l}{x}^{2}-5x+6=0\\ \left(3{\right)}^{2}-5\left(3\right)+6=0\\ 0=0\end{array}$
x = 3 satisfies the equation.

Conclusion:
1. 2 and 3 satisfy the equation ${x}^{2}-5x+6=0$ , hence there are the roots of the equation.
2. 1 does not satisfy the equation ${x}^{2}-5x+6=0$ , hence it is NOT the root of the equation.
Example 2:
Solve the quadratic equation  $\frac{2{x}^{2}-5}{3}=3x$

Solution:
$\frac{2{x}^{2}-5}{3}=3x$
2x2 – 5 = 9x
2x2 – 9x – 5 = 0
(x – 5)(2x + 1) = 0
x – 5 = 0, x = 5
or 2x + 1 = 0
Therefore, = 5 and x = ½ are roots or solutions of the quadratic equation.

Example 3:
Solve the quadratic equation 4x2 – 12 = –13x

Solution:
4x2 – 12 = –13x
4x2 + 13x – 12 = 0
(4x – 3)(x + 4) = 0
4x – 3 = 0,  $x=\frac{3}{4}$
or x + 4 = 0
x = –4

Example 4
:
Solve the quadratic equation 5x2 = 3 (x + 2) – 4

Solution:
5x2 = 3 (x + 2) – 4
5x2 = 3x + 6 – 4
5x2 – 3x – 2 = 0
(5x + 2)(x – 1) = 0
5x + 2 = 0,  $x=-\frac{2}{5}$
or x – 1 = 0
x = 1

Example 5:
Solve the quadratic equation
$\frac{3x\left(x-3\right)}{4}=-x+3.$

Solution:
$\frac{3x\left(x-3\right)}{4}=-x+3$
3x2 – 9x = – 4x + 12
3x2 – 5x – 12 = 0
(3x + 4)(x – 3) = 0
3x + 4 = 0,  $x=-\frac{4}{3}$
or x – 3 = 0
x = 3