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. This method can be used to solve and factorise quadratics where previous methods could not, and is in fact used to prove the quadratic formula in the next lesson. This method will also be very useful in helping to identify key features in graphs such as the turning point of a quadratic or centre of a circle.
Visually
Consider the following quadratic expression:
$x^2+4x-5$x2+4x−5
This expression can be factorised using the sum and product method, but let's practice 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
- Use difference of two squares to write the final factorisation
Once the quadratic has been factorised, the null factor law or even algebraic manipulation before the final step of factorisation, could be used to find solutions.
Worked example
Question 1
a) Factorise $x^2+4x-5$x2+4x−5 by first completing the square.
Think: Here is a visual of the method:
From the video, we can see that $x^2+4x-5=\left(x+2\right)^2-9$x2+4x−5=(x+2)2−9.
Do: To factorise the expression, use the difference of two squares:
| $x^2+4x-5$x2+4x−5 | $=$= | $\left(x+2\right)^2-9$(x+2)2−9 |
| $=$= | $\left(x+2+3\right)\left(x+2-3\right)$(x+2+3)(x+2−3) | |
| $=$= | $\left(x+5\right)\left(x-1\right)$(x+5)(x−1) |
b) Hence, solve the equation $x^2+4x-5=0$x2+4x−5=0:
Do: We can now solve the equation easily with the null factor law.
| $x^2+4x-5$x2+4x−5 | $=$= | $0$0 |
| $\left(x+5\right)\left(x-1\right)$(x+5)(x−1) | $=$= | $0$0 |
So either
| $x+5$x+5 | $=$= | $0$0 | or | $x-1$x−1 | $=$= | $0$0 |
| $x$x | $=$= | $-5$−5 | $x$x | $=$= | $1$1 |
Note: If we were only asked to solve the equation, we could stop before the factorisation and solve using algebraic manipulation as follows:
| $x^2+4x-5$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 the solutions $x=1$x=1 or $x=-5$x=−5.
This interactive will help you to visualise the process.
Watch this video for an explanation .
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Algebraically
Let's now look at how we can perform this method using algebraic steps.
Factorising a quadratic expression algebraically:
- Factor out the coefficient of the $x^2$x2 term, unless it is $1$1
- Halve the coefficient of $x$x and square it, add and subtract this value
- Rewrite the terms that now form a perfect square
- Combine the constant and term left over from completing the square
- Use difference of two squares to fully factorise
Solving a quadratic equation algebraically:
- Divide both sides of the equation by the coefficient of the $x^2$x2 term
- Move the constant term to the right-hand side of the equation
- Halve the coefficient of the $x$x term and square it, add this value to both sides of the equation
- Rewrite the left-hand side of the equation as a perfect square
- Rearrange using algebraic manipulation
Practice questions
Question 2
Using the method of completing the square, rewrite $x^2+3x+6$x2+3x+6 in the form $\left(x+b\right)^2+c$(x+b)2+c.
Question 3
Factorise the quadratic using the method of completing the square to get it into the form $y=\left(x+a+\sqrt{b}\right)\left(x+a-\sqrt{b}\right)$y=(x+a+√b)(x+a−√b).
$y=x^2+8x+14$y=x2+8x+14
Question 4
Solve $x^2-6x-16=0$x2−6x−16=0 by completing the square:
Question 5
Solve the following quadratic equation by completing the square:
$4x^2+11x+7=0$4x2+11x+7=0
Write all solutions on the same line, separated by commas.
Enter each line of work as an equation.
Question 6
Solve for $x$x by first completing the square. Express your solutions in simplest fraction form. $x^2-9x+2=0$x2−9x+2=0