METHODS FOR SOLVING QUADRATIC EQUATION

There are 3 basic methods to solve an Quadratic Equation: –

1. By Quadratic Formula
2. By Splitting the middle term
3. By completing the square

In this session ,we will discuss how we can solve an equation by Quadratic formula:)

By Quadratic Formula

As we discussed this before in the concept “ Basics of quadratic equations”. The Quadratic Equation Formula is: –

(α, β) = [-b ± √(b² – 4ac)]/2a  or we can write it as [-b ± √D]/2a As: –

D = b² – 4ac

Now, Let us understand the application of quadratic formula by finding the roots of some Quadratic Equation:

( 1 ). X²  + 2x – 35  =  0

According to the general quadratic form: ax² + bx +c = 0

Here , a = 1 , b = 2, c = -35

You can follow the simple following steps to get your roots: –

Step 1 = Firstly, find the discriminant using b² – 4ac  (to put in the quadratic formula): –

D = b² – 4ac

= (2)2 – 4 x 1 x -35
= 4 + 140 = 144

Our discriminant is 144.

Step 2 = Now, put the discriminant in our quadratic formula.

x = [-b ± √(b² – 4ac)]/2a

= -2 ± √144 / 2*1
= -2 + 12 /2                                     = -2-12/2
= 10/2                                              = -14/2
= 5                                                     = -7

So, the final roots of the quadratic equation are 5 and -7.

( 2 ). 2x²-6x+3 = 0

Here , a = 2 , b = -6, c = 3

Step 1 = Firstly, find the discriminant using b² – 4ac  (to put in the quadratic formula): –
D = b2 – 4ac
= (-6)2 – 4 x 2 x 3
= 36 – 24 = 12

Step 2 = Now, put the discriminant in our quadratic formula.

x = [-b ± √(b² – 4ac)]/2a

= -(-6) ± √12 / 2x 2
= 6 + √12 /4                                    = 6 – √12 /4

So, the final roots of the Quadratic Equation are 6 + √12 /4 and 6 – √12 /4.

( 3 ) .  x²+5x+4 = 0

Here , a = 1 , b = 5, c = 4

Step 1 = Firstly, find the discriminant using b² – 4ac (to put in the quadratic formula): –

D = b² – 4ac

= (5)2 – 4 x 1 x 4
= 25 – 16 = 9

Step 2 = Now, put the discriminant in our quadratic formula.

x = [-b ± √(b² – 4ac)]/2a

= -5 ± √9 / 2 x 1
=  -5 + 3 /2 = -5 – 3 / 2
= -2/2 = -8/2
= -1 = -4

So, the final roots of the quadratic equation are -1 and -4.

( 4 )  .  x² +10x + 9 = 0

Here , a = 1 , b = 10, c = 9

Step 1 = Firstly, find the discriminant using b2 – 4ac (to put in the quadratic formula): –
D = b² – 4ac

(10)² – 4 x 1 x 9

= 100 – 36 = 64

Step 2 = Now, put the discriminant in our quadratic formula.

x = [-b ± √(b² – 4ac)]/2a

= -10 ± √64 / 2 x 1
= -10 + 8 /2 = -10 – 8/2
= -2/2 = -18 / 2
= -1 = -9

So, the final roots of the quadratic equation are -1 and -9.

( 5 ) .  6x² -x – 5 = 0

Here , a = 6 , b = -1, c = -5

Step 1 = Firstly, find the discriminant using b2 – 4ac (to put in the quadratic formula): –
D = b² – 4ac

= (-1)² – 4 x 6 x -5

= 1 + 120 = 121

Step 2 = Now, put the discriminant in our quadratic formula.

x = [-b ± √(b² – 4ac)]/2a

= -(-1) ± √121 / 2 x 6
= 1 + 11 /12 = 1-11/12
= 12/12 = -10/12
= 1 = -5/6

So, the final roots of the quadratic equation are 1 and -5/6.

6.  12x² -25x = 0

Here , a = 12 , b = -25, c = 0

Step 1 = Firstly, find the discriminant using b2 – 4ac (to put in the quadratic formula): –
D = b² – 4ac

= (-25)² – 4 x 12 x 0

= 625-0 = 625

Step 2 = Now, put the discriminant in our quadratic formula.

x = [-b ± √(b² – 4ac)]/2a

= -(-25) ± √625 / 2 x 12
= 25 + 25 /24 = 25-25/24
= 50/24 = 0/24
= 25/12 = 0

So, the final roots of the quadratic equation are 25/12 and 0.

I hope now, you can easily understand the topic “ solving a quadratic equation by quadratic formula”.

Read More : TRIANGLE: Definition, Parts, Properties,Types and Formulae

Lets Practice

• 3x² + x + 1 = 0
• x² + 4x – 5 = 0
• 2x² + x – 528 = 0
•  x² -11x = -28
• x² – 4 = 0