Solve a Quadratic Equation by Factoring

A quadratic equation is a polynomial equation with degree two.Its general form is

$$ax^2 + bx +c = 0$$

Roots of quadratic equation

As the degree of quadratic equation is 2 so the number of roots is also 2. That is there are 2 values of x

for which the equation $$ax^2 + bx +c = 0$$ is satisfied.

Solving by Factorization

for a quadratic equation ,$$ax^2 + bx +c = 0$$ , the sum of roots is -b/a and product of roots is c/a.

so we try to split the middle term bx such that it satisfies both these conditions.

And finally we have to covert it in the form (x+ m) ( x+ n).

If the expression is in product form (x + ..) (x+.. ) = 0

then either of the terms can be zero so that product is zero.

put both the terms = 0 and we get the roots

x+m = 0 so,x = -m

x+n = 0 so, x = -n

Examples:

• $$ x^2 + 7x + 12 = 0$$

splitting 7x = 4x + 3x

as 4 X3 = 12 and 4 + 3 = 7

so we have

$$x^2 + 4x + 3x + 12$$

=$$x(x+ 4) + 3( x+ 4)$$

=> (x+ 4)( x+ 3) = 0

now either of the terms can be zero

putting (x+ 4) = 0

so, x = - 4

or (x+ 3) = 0

so, we get x = -3

Roots of the equation are x = -3, -4

• $$2x^2 + 5x - 25 = 0$$

now (2) (-25) = -50

so , splitting 5x as 10x - 5x

$$2x^2 + 10x - 5x -25 $$

= 2x(x+5) -5(x+5)

=> (x+5)(2x-5) = 0

so , either (x+5) = 0 , we get x = -5

or (2x -5) = 0 , we get x = 5/2

Roots for given quadratic equation are x = -5,5/2