How to solve quadratic equations by using Quadratic equation formula|Discriminant of a quadratic equation

3)Solving quadratic equations by using Quadratic formula

For quadratic equation

ax2 +bx+c=0,

roots are given by

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

and

$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

This has been derived using the square method as discussed above only

For b2 -4ac  > 0, Quadratic equation has two real roots of different value

For b2-4ac =0, quadratic equation has one real root

For b2-4ac < 0, no real roots for quadratic equation

Example

3x2 - 5x + 2 = 0

step 1 Here a =3 ,b=-5 and c=2
step 2 Substituting in these formula
$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

and

$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

step 3 we get the root as 2/3 and 1

Nature of roots of Quadratic equation

b2 -4ac is called the discriminant of the quadratic equation
 S.no Condition Nature of roots 1 b2 -4ac  > 0 Two distinct real roots 2 b2-4ac =0 One real root 3 b2-4ac < 0 No real roots

Problem based on discriminant of a quadratic equation

Example

The sum of the ages of two friends A and B is 20 years. Four years ago, the product of their ages in years was 48.

Here we need to advise if that is a possible condition

Let the age of A be x years.
then the age of the B will be (20 - x) years.
Now 4 years ago,
Age of A = (x - 4) years
Age of N = (20 - x - 4) = (16 - x) years
So we get that,
(x - 4) (16 - x) = 48
16x - x2 - 64 + 4x = 48
x2 - 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b2 - 4ac
= (-20)2 - 4 * 112
= 400 - 448 = -48

b2 - 4ac < 0 Therefore, there will be no real solution possible for the equations. Such type of condition doesn't exist

Example

Is the following statement 'True' or 'False'?Justify your answer.
If in a quadratic equation the coefficient of x is zero, then the quadratic equation has no
real roots

False, since the discriminant in this case is -4ac which can still be nonnegative if a and c are of opposite signs or if one of a or c is zero.

Solved examples

Example-1
Find the roots of the quadratic equation x2 -6x=0

Solution

There is no constant term in this quadratic equation, we can x as common factor

x(x-6)=0

So roots are x=0 and x=6

Example-2
Solve the quadratic equation x2  -16=0

Solution

x2 -16=0

x2 =16

or x=4 or -4

Example-3
Solve the quadratic equation by factorization method x2 -x -20=0

Solution

Step 1 First we need to multiple the coefficient a and c.In this case =1X-20=-20. The possible multiple are 4,5 ,2,10

Step 2 The multiple 4,5 suite the equation

x2-5x+4x-20=0

x(x-5)+4(x-5)=0

(x+4)(x-5)=0

or x=-4 or 5

Example-4
Solve the quadratic equation by Quadratic method x2 -3x-18=0

Solution

For quadratic equation

ax2 +bx+c=0,

roots are given by

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

and

$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

Here a=1

b=-3

c=-18

Substituting these values,we get

x=6  and -3