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Stage 5.1-3

2.13 Completing the square

Lesson

Completing the square

Earlier we saw the identity \left(A+B\right)^2=A^2+2AB+B^2. Sometimes it's useful to be able to have a perfect square. However, not every quadratic trinomial can be factorised this way.

Instead we can use a method called completing the square. The idea behind completing the square is that some part of the original expression is a perfect square, and the remainder is a constant.

Complete the square on x^2-4x+23.

First, notice that x^2-4x+23 is not a perfect square, because it doesn't fit the pattern A^2+2AB+B^2. Instead we want to find out which part does fit this pattern.

The first term is x^2, and so we want A=x. Substituting this in gives us x^2+2Bx+B^2. So what is B? We can see that -4x=2Bx and from this we can solve for B.

\displaystyle -4x\displaystyle =\displaystyle 2BxEquating the x terms
\displaystyle -4\displaystyle =\displaystyle 2BDivide both sides by x
\displaystyle -2\displaystyle =\displaystyle BDivide both sides by 2

So B=-2. Substituting this gives us A^2+2AB+B^2=x^2-4x+4. While this is not quite the expression we started with, if we can separate this from the expression then we have a perfect square.

If we add 4 and then subtract 4 from a number then we end up with the number that we started with. In other words, +4-4 is the same as 0. Applying this to the expression:

\displaystyle x^2-4x+23\displaystyle =\displaystyle x^2-4x+4-4+23Adding 4-4 into the equation
\displaystyle =\displaystyle \left(x-2\right)^2-4+23Factorise the perfect square part
\displaystyle =\displaystyle \left(x-2\right)^2+19Evaluate the constant

And now we have a perfect square plus a constant. In generality, let A, B, and C be any numbers. Applying the whole process:

\displaystyle A^2+2AB+C\displaystyle =\displaystyle A^2+2AB+B^2-B^2+CAdding B^2-B^2 into the equation
\displaystyle =\displaystyle \left(A+B\right)^2-B^2+CFactorise the perfect square part

So we can complete the square on any expression. We find A by finding the squared term, and then we find B by dividing the middle term by 2A. Then we add B^2-B^2 and we have a perfect square plus a constant.

Examples

Example 1

Using the method of completing the square, rewrite x^2+4x in the form \left(x+b\right)^2+c

Worked Solution
Create a strategy

Add b^2-b^2 into the expression and factorise the perfect square part.

Apply the idea

If x^2+2bx=x^2+4x then 2b=4 so b=2.

\displaystyle x^2+4x\displaystyle =\displaystyle x^2+4x +2^2-2^2Adding 2^2-2^2
\displaystyle =\displaystyle (x+2)^2-2^2Factorise the perfect square part
\displaystyle =\displaystyle (x+2)^2-4Evaluate the constant

Example 2

Factorise the quadratic y=x^2+4x+3 using the method of completing the square to get it into the form y=\left(x+a\right)\left(x+b\right).

Worked Solution
Create a strategy

Use the completing the square method.

Apply the idea

We have a coefficient of 4 in the x term, then 2B=4 so B=2.

\displaystyle y\displaystyle =\displaystyle x^2+4x+2^2-2^2+3Add 2^2-2^2
\displaystyle =\displaystyle \left(x+2\right)^2-2^2+3Factorise the perfect square part
\displaystyle =\displaystyle \left(x+2\right)^2-1Evaluate the constant
\displaystyle =\displaystyle \left(x+2\right)^2-1^2Write as difference of 2 squares
\displaystyle =\displaystyle \left((x+2)+1\right)\left((x+2)-1\right)Factorise by difference of 2 squares
\displaystyle =\displaystyle \left(x+3\right)\left(x+1\right)Simplify each bracket
Idea summary

Completing the square is a method which allows us to find a perfect square by using the rule \left(A+B\right)^2=A^2+2AB+B^2. The method involves the following steps:

  1. Find A by taking the square root of the squared term.

  2. Find B by dividing the middle term by 2A.

  3. Add B^2-B^2 to the expression (which is the same as adding 0).

  4. Factorise the perfect square part of the expression using the rule\left(A+B\right)^2=A^2+2AB+B^2.

  5. Collect the constant terms if necessary.

Outcomes

MA5.2-8NA

solves linear and simple quadratic equations, linear inequalities and linear simultaneous equations, using analytical and graphical techniques

MA5.3-5NA

selects and applies appropriate algebraic techniques to operate with algebraic expressions

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