In your journey of factorising quadratics, you will realise that some quadratic expressions are difficult to factorise using the normal method of multiplying two brackets. Take for example, \(x^2+x+1\). You will not be able to factorise this. This is where completing a square comes in.
Completing a square is a method where we add and subtract the square of half the coefficient of \(x\).
The technique of completing a square will be more beneficial to you when you take convert a quadratic from a standard form, \(\normalsize y=ax^2+bx+c\), to \(\normalsize y=a(x+p)^2+q\)
In this article you will learn to simplify quadratic expressions by completing a square and also how to solve quadratic equations by completing a square.
Simplifying an expression by completing a square.
To simplify an expression by completing a square, do the following
- Factor out the coefficient of \(\normalsize x^2\) so that it becomes 1.
- Add and subtract the square of half the coefficient of \(\normalsize x\)
- The first three terms are the perfect-square trinomial. Refer to the example below
- Simplify the expression by adding the constants
- Multiply the simplified terms with the factored number.
Example: Simplify the following expressions by completing a square.
$$ \normalsize \begin{align*} &i. y=x^2+x+1\\ &ii. y=3x^2+7x-6 \end{align*} $$ Solutions-
\(\normalsize a=1\), so there is no need to factor out.
$$ \normalsize \begin{align*} y&=x^2+x+1\\ &= x^2 +x+ \left( \frac{1}{2} \times 1 \right) ^{2} - \left( \frac{1}{2} \times 1 \right) ^{2}+1 \\ &= x^2 + x+ \left( \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ &= \left( x+ \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ &= \left( x+ \frac{1}{2} \right) ^{2} + \frac{3}{4} \end{align*} $$The perfect-square trinomial was \(\normalsize x^2+x+(\frac{1}{2})^2\)
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\(\normalsize a=3\), so we need to factor out by 3.
$$ \normalsize \begin{align*} y&=3x^2+7x-6\\ &=3 \left( x^2 + \normalsize \frac{7x}{3} - 2 \right) \\ &= 3 \left[ x^2 + \frac{7x}{3}+ \left( \frac{1}{2} \times \frac{7}{3} \right) ^{2} - \left( \frac{1}{2} \times \frac{7}{3} \right) ^{2}-2 \right] \\ &= 3 \left[ x^2 + \frac{7x}{3}+ \left( \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ &= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ &= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{36} \right]\\ &= 3 \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{12} \end{align*} $$
Solving for \(x\) by completing a square.
The procedure is a bit different when we solve for \(x\). However, we are almost doing the same things.
The method:
- Divide throughout by the coefficient of \(x^2\).
- Take the constant to the right hand side of the equation.
- The above steps can be done at all once.
- Add the square of half the coefficient of \(x\) to both sides of the equation.
- Simplify LHS which is the perfect-square trinomial and simplify the RHS too.
- Square root both sides and prefix the RHS with a \(\pm\).
- Now solve for \(x\).
Example: Solve for \(x\) by completing a square.
$$ \normalsize \begin{align*} &i. x^2+x+1=0\\ &ii. 3x^2+7x-6=0 \end{align*} $$Solutions
\(\normalsize x^2+x+1=0\)
$$ \normalsize \begin{align*} x^2 +x+ \left( \frac{1}{2} \times 1 \right) ^{2} &= \left( \frac{1}{2} \times 1 \right) ^{2}-1 \\ x^2 + x+ \left( \frac{1}{2} \right) ^{2} &= \left( \frac{1}{2} \right) ^{2}-1 \\ \left( x+ \frac{1}{2} \right) ^{2} &= - \frac{3}{4}\\ \sqrt{\left( x+ \frac{1}{2} \right) ^{2}} &= \pm \sqrt{ \frac{-3}{4}}\\ x+ \frac{1}{2} &= \pm \frac{{\sqrt{-3}}}{2}\\ x&=- \frac{1}{2} \pm \frac{{\sqrt{-3}}}{2}\\ x &= \frac{{-1 \pm \sqrt{-3}}}{2} \end{align*} $$The solution to this expression is non-real. This is beyond our scope. So we leave it here.
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\(\normalsize 3x^2+7x-6=0\)
$$ \normalsize \begin{align*} 3x^2+7x&=6\\ x^2+\frac{7}{3}{x}&=2\\ x^2 + \frac{7x}{3}+ \left( \frac{1}{2} \times \frac{7}{3} \right) ^{2} &= \left( \frac{1}{2} \times \frac{7}{3} \right) ^{2}+2\\ x^2 + \frac{7x}{3}+ \left( \frac{7}{6} \right) ^{2} &= \left( \frac{7}{6} \right) ^{2}+2 \\ \left( x+ \frac{7}{6} \right) ^{2} &= \frac{121}{36} \\ \sqrt{\left( x+ \frac{7}{6} \right) ^{2}} &= \pm \sqrt{\frac{121}{36}}\\ x+ \frac{7}{6} &= \pm \frac{11}{6} \\ x &=- \frac{7}{6} \pm \frac{11}{6} \\ x = \frac{2}{3} &\textrm{ or } x = -3 \end{align*} $$Once you get used to this procedure, the steps can be shortened. Ours are long because we are trying to show you how to complete a square with in detail.