# Solve by completing the square I

Lesson

Completing the square is the process of rewriting a quadratic expression from the expanded form $x^2+2bx+c$x2+2bx+c into the square form $\left(x+b\right)^2+c-b^2$(x+b)2+cb2. The name comes from how the method was first performed by ancient Greek mathematicians who were actually looking for numbers to complete a physical square.

There are two ways to think about the completing the square method: visually (using a square), and algebraically. We will see that we can use this method to find the solutions to the quadratic equation and to more easily graph the function.

## Visually

Consider the following quadratics:

$x^2+4x-5$x2+4x5

$x^2+2x-3$x2+2x3

$x^2-8x-20$x28x20

Let's imagine we want to factorise them using the complete the square method.

The steps involved are:

• Move the constant term to the end
• Create a box and complete the square
• Keep the expression balanced
• Write the factorisation

As this is a visual method, its best to learn by watching the process in action, so here are three examples.

##### Question 1

We wish to solve $x^2+4x-5=0$x2+4x5=0 by first completing the square.

Completing the square on $x^2+4x-5$x2+4x5:

Then we solve this algebraically.

 $x^2+4x-5$x2+4x−5 $=$= $0$0 $\left(x+2\right)^2-9$(x+2)2−9 $=$= $0$0 $\left(x+2\right)^2$(x+2)2 $=$= $9$9 $x+2$x+2 $=$= $\pm3$±3

Then by subtracting $2$2 from both sides of the equation we have $x=1$x=1 or $x=-5$x=5.

##### Question 2

We wish to solve $x^2+2x-3=0$x2+2x3=0 by first completing the square.

Completing the square on $x^2+2x-3$x2+2x3:

Then we solve this algebraically.

 $x^2+2x-3$x2+2x−3 $=$= $0$0 $\left(x+1\right)^2-4$(x+1)2−4 $=$= $0$0 $\left(x+1\right)^2$(x+1)2 $=$= $4$4 $x+1$x+1 $=$= $\pm2$±2

And by subtracting $1$1 from both sides of the equation we have $x=1$x=1 or $x=-3$x=3.

##### Question 3

We wish to solve $x^2-8x-20=0$x28x20=0 by first completing the square.

Completing the square on $x^2-8x-20$x28x20:

Then we solve this algebraically.

 $x^2-8x-20$x2−8x−20 $=$= $0$0 $\left(x-4\right)^2-36$(x−4)2−36 $=$= $0$0 $\left(x-4\right)^2$(x−4)2 $=$= $36$36 $x-4$x−4 $=$= $\pm6$±6

We can add 4 to both sides of the equation to get $x=10$x=10 or $x=-2$x=2.

This interactive will help you to visualise the process. Watch this video for an explanation .

## Algebraically

Let's imagine we want to factorise the same quadratics as before, but this time using the algebraic version of the complete the square method.

The steps involved are:

• Move the constant term to the end
• Add half the coefficient of $x$x and square it
• Keep the expression balanced
• Complete the square

#### Worked Examples

##### Question 1

Solve the following quadratic equation by completing the square:

$x^2+18x+32=0$x2+18x+32=0

##### Question 2

Solve $x^2-6x-16=0$x26x16=0 by completing the square:

##### Question 3

Solve for $x$x by first completing the square.

$x^2-2x-32=0$x22x32=0

##### Question 4

Solve the following quadratic equation by completing the square:

$4x^2+11x+7=0$4x2+11x+7=0

1. Write all solutions on the same line, separated by commas.