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How to solve quadratic equations by using Quadratic equation formula




3)Solving quadratic equations by using Quadratic formula

For quadratic equation

ax2 +bx+c=0,
roots are given by
quadratic Formula
$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$
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
Solution
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.
Solution
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
Solution
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





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