Solving Quadratic Equations

Introduction :

An equation in the form of ax2 + bx + c = 0, where a, b, c ? R and a ?0 is called a quadratic equation. It is an equation with degree 2. For example, 5x2 + 6x + 7 = 0, 9x2 – 4 = 0, 2x2 – 3x =0 are quadratic equations. If p(x) = 0 is a quadratic equations, then the zero of the polynomial  p(x) are called the roots of equation p(x) = 0. Find the roots of a quadratic equation is called as solve the quadratic equations.

Solution of Quadratic Equations Example

To solve quadratic equation by using factorization method using at what time the quadratic equation is expressible as the product of two linear equations. Quadratic equation of the form ax2+bx+c=0 here a,b,c,d are called constants can be used to reduced the form of a quadratic equations.

Example 1:

Solving quadratic equations X2+9x-10

Solution:

Step 1:

List out the factors

1x-10, -1x10

Step 2:

Find the factor the sum value is 9

-1+10=9

Step 3:

Check using distributive property

(X-1)(x+10)=x2+10x-x-10

X2+9x-10

Step 4: back to original equation

X2+9x-10 we get two values for x.

X-1=0=>x=1

x+10=0=>x=-10

Answer:  x=1, x=-10

Example 2:

Solving the quadratic equations 2x2 + 3x – 5 = 0

Solution: 2x2 + 3x – 5 = 2x2 + 5x – 2x – 5

= x (2x + 5) – 1(2x + 5)

= (x – 1) (2x + 5)

Since 2x2 + 3x – 5 = 0 we get (x – 1) (2x + 5) = 0

=>x – 1 = 0 or 2x + 5 = 0

=>x = 1, –5/2

The solution set = 1, -5/2

Example 3:

Solving the quadratic equations 64x2 – 36 = 0

Solution: 64x2 – 36 = 0 or (8x + 6) (8x – 6) = 0

Or x = –6/8 = –3/4

Or x = 6/8 = 3/4

The solution set =-3/4, 3 /4

Example 4:

Solving the quadratic equations 3x2 – 4x = 0

Solution: 3x2 – 4x = x (3x – 4)

Since 3x2 – 4x = 0, we get x (3x – 4) = 0

Or x = 0

Or 3 x –4 = 0

=> 3x = 4

=> x = 4/3

Quadratic Equation Practice Problem

1) Solving the quadratic equations x2+4x-5=0

Answer:

X=1, x=-5

2) Solving the quadratic equations x2-5x-6

Answer:

x=-1, x=6