Solving quadratic equations by using Quadratic formula
For quadratic equation
$ax^2 +bx+c=0$,
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 below
$ax^2 +bx+c=0$
$x^2 +\frac {b}{a} x+\frac {c}{a}=0$
$(x+ \frac {b}{2a})^2 - (\frac {b}{2a})^2 + \frac {c}{a}=0$
$(x+ \frac {b}{2a})^2=\frac {(b^2-4ac)}{4a^2}$
$(x+ \frac {b}{2a})= \pm \frac {\sqrt {b^2-4ac}}{2a}$
$x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$
For $b^2 -4ac > 0$, Quadratic equation has two real roots of different value
For $b^2 -4ac =0$, quadratic equation has one real root
For $b^2 -4ac < 0$, no real roots for quadratic equation
Example
$3x^2 - 5x + 2 = 0$ Solutionstep 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
$b^2 -4ac $ is called the discriminant of the quadratic equation
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,
$(x - 4) (16 - x) = 48$
$16x - x^2 - 64 + 4x = 48$
$x^2 - 20x + 112 = 0$
Comparing this equation with $ax^2 + bx + c = 0$, we get
a = 1, b = -20 and c = 112
Discriminant = $b^2 - 4ac$
= (-20)2 - 4 * 112
= 400 - 448 = -48
$b^2 - 4ac <0$
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 non-negative
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 $x^2 -6x=0$ 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 $x^2 -16=0$ Solution
$x^2 -16=0$
$x^2 =16$
or$x=4 \; or \; -4$
Example-3 Solve the quadratic equation by factorization method $x^2 -x -20=0$ 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
$x^2-5x+4x-20=0$
$x(x-5)+4(x-5)=0$
$(x+4)(x-5)=0$
or $x=-4 \; or \; 5$
Example-4 Solve the quadratic equation by Quadratic method $x^2 -3x-18=0$ Solution
For quadratic equation
$ax^2 +bx+c=0$,
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
$x=6 \; and \; -3$
