Solving by completing the square I

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+c−b2. 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+4x−5

$x^2+2x-3$x2+2x−3

$x^2-8x-20$x2−8x−20

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+4x−5=0 by first completing the square.

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

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+2x−3=0 by first completing the square.

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

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 $x^2-8x-20=0$x2−8x−20=0 by first completing the square.

Completing the square on $x^2-8x-20$x2−8x−20:

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 Watch video.

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

Question 2

Question 3

Question 4