1. Solving Quadratic Equations by Factoring

The general form of a quadratic equation is

ax2 + bx + c = 0

where x is the variable and a, b & c are constants

(a) 5x2 − 3x − 1 = 0 is a quadratic equation in quadratic form where

a = 5, b = -3, c = -1

(b) 5 + 3t − 4.9t2 = 0 is a quadratic equation in quadratic form.

Here, a = -4.9, b = 3, c = 5

[This equation arose from finding the time when a projectile, being acted on by gravity, hits the ground.]

(c) (x + 1)2 = 4 is a quadratic equation but not in quadratic form.

It has to be expanded and simplified to:

x2 + 2x − 3 = 0

Continues below

Summary

• must contain an x2 term
• must NOT contain terms with degrees higher than x2 eg. x3, x4 etc

• bx − 6 = 0 is NOT a quadratic equation because there is no x2 term.
• x3x2 − 5 = 0 is NOT a quadratic equation because there is an x3 term (not allowed in quadratic equations).

The solution of an equation consists of all numbers (roots) which make the equation true.

All quadratic equations have 2 solutions (ie. 2 roots). They can be:

• real and distinct
• real and equal
• imaginary (complex)

Example 1

The quadratic equation x2 − 7x + 10 = 0 has roots of

x = 2 and x = 5. (We'll show below how to find these roots.)

This can be seen by substituting in the equation:

When x = 2,

x2 − 7x + 10

= (2)2 − 7(2) + 10

= 4 − 14 + 10

= 0

(This can be shown similarly for x = 5). In this example, the roots are real and distinct.

Example 2

The quadratic equation x2 − 6 x + 9 = 0 has double roots of x = 3 (both roots are the same)

This can be seen by substituting x = 3 in the equation:

x2 − 6x + 9

= (3)2 − 6(3) + 9

= 9 − 18 + 9

= 0

Example 3

x2 + 9 = 0

has imaginary roots of

x=sqrt(-9) or -sqrt(-9)

Solving a Quadratic Equation by Factoring

For the time being, we shall deal only with quadratic equations that can be factored (factorised). If you need a reminder on how to factor, go back to the section on Factoring Trinomials.

Using the fact that a product is zero if any of its factors is zero we follow these steps:

(i) Bring all terms to the left and simplify, leaving zero on the right side.

(iii) Set each factor equal to zero

(iv) Solve the resulting linear equations

(v) Check the solutions in the original equation

Example 4

Solve x2 − 2x − 15 = 0

x2 − 2x − 15 = 0

Factoring gives:

(x − 5)(x + 3) = 0

Now, if either of the terms (x − 5) or (x + 3) is 0, the product is zero. So we conclude:

(x − 5) = 0, therefore

x = 5

or

(x + 3) = 0, therefore

x = − 3

Hence the roots are x = 5 and x = − 3.

Are we correct?

We check the roots in the original equation by substitution.

When x = 5:

x2 − 2x − 15

= (5)2 − 10 − 15

= 25 − 10 − 15

= 0

(Similarly, when we substitute x = -3, we also get 0.)

Example 5

Solve

9x^2+ 6x + 1 = 0

9x2 + 6x + 1 = 0

Factoring gives:

(3x + 1)(3x + 1) = 0

So we conclude:

(3x + 1) = 0,

therefore

x=-1/3

We say there is a double root of x = -1/3.

Example 6 (involving fractions)

Solve

2-1/x=3/(x+2)

2-1/x=3/(x+2)

Multiply throughout by x(x+2) to remove the denominators (bottoms) of the fractions:

2x(x+2)-(x(x+2))/x=(3(x)(x+2))/(x+2)

Cancelling gives:

2x(x+2)-(x+2)=3x

Expanding the brackets:

2x^2+4x-x-2=3x

2x^2-2=0

x^2-1=0

Factoring gives:

(x+1)(x-1)=0

So x = -1 or x = 1.

CHECK: Substituting x = -1 into both the left hand side and right hand side of the question gives:

"LHS"=2-1/x=2-1/-1=3

"RHS"=3/(x+2)=3/(-1+2)=3="LHS"

Likewise, for x = +1,

LHS = 2 - 1 = 1

RHS = 3/3 = 1 = LHS

Exercises

1. Determine if the following are quadratic equations. If so, determine a, b, and c.

a. 5x2 = 9 − x

b. (3x − 2)2 = 2

Q1a    5x^2= 9 − x

5x^2+ x − 9 = 0

So, yes, it is a quadratic equation with

a = 5, b = 1, c = -9

Q1b    (3x − 2)^2= 2

9x^2− 12x + 4 = 2

9x^2− 12x + 2 = 0

So, yes, it is a quadratic equation with

a = 9, b = -12, c = 2

Get the Daily Math Tweet!

1. Solve for x:

2x2 − 7x + 6 = 3

x=1/2 or x=3.