How to Complete a Square.
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, \(y=ax^2+bx+c\), to
$${\small y= a (\, x+p )\,^{2}+q}$$Simplifying an expression by completing a square.
To simplify an expression by completing a square, do the following
- Factor out the coefficient of \(x^2\) so that it becomes 1.
- Add and subtract the square of half the coefficient of \(x\)
- 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.
\begin{gather*} \small i. y=x^2+x+1\\ \small ii. y=3x^2+7x-6 \end{gather*}-
\(a=1\), so there is no need to factor out.
\begin{gather*} \small y=x^2+x+1\\ \small y= x^2 +x+ \left( \frac{1}{2} \times 1 \right) ^{2} - \left( \frac{1}{2} \times 1 \right) ^{2}+1 \\ \small y= x^2 + x+ \left( \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ \small y= \left( x+ \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ \small y= \left( x+ \frac{1}{2} \right) ^{2} + \frac{3}{4} \end{gather*}The perfect-square trinomial was \(x^2+x+(\frac{1}{2})^2\)
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\(a=3\), so we need to factor out by 3.
\begin{gather*} \small y=3x^2+7x-6\\ \small y=3 \left( x^2 + \small \frac{7x}{3} - 2 \right) \\ \small y= 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] \\ \small y= 3 \left[ x^2 + \frac{7x}{3}+ \left( \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ \small y= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ \small y= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{36} \right]\\ \small y= 3 \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{12} \end{gather*}
Solutions
Example: Simplify the following expressions by completing a square.
\begin{gather*} \small i. y=x^2+x+1\\ \small ii. y=3x^2+7x-6 \end{gather*}-
\(a=1\), so there is no need to factor out.
\begin{gather*} \small y=x^2+x+1\\ \small y= x^2 +x+ \left( \frac{1}{2} \times 1 \right) ^{2} - \left( \frac{1}{2} \times 1 \right) ^{2}+1 \\ \small y= x^2 + x+ \left( \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ \small y= \left( x+ \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ \small y= \left( x+ \frac{1}{2} \right) ^{2} + \frac{3}{4} \end{gather*}The perfect-square trinomial was \(x^2+x+(\frac{1}{2})^2\)
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\(a=3\), so we need to factor out by 3.
\begin{gather*} \small y=3x^2+7x-6\\ \small y=3 \left( x^2 + \frac{7x}{3} - 2 \right) \\ \small y= 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] \\ \small y= 3 \left[ x^2 + \frac{7x}{3}+ \left( \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ \small y= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ \small y= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{36} \right]\\ \small y= 3 \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{12} \end{gather*}
Solutions
Example: Simplify the following expressions by completing a square.
\begin{gather*} \small i. y=x^2+x+1\\ \small ii. y=3x^2+7x-6 \end{gather*}-
\(a=1\), so there is no need to factor out.
\begin{gather*} \small y=x^2+x+1\\ \small y= x^2 +x+ \left( \frac{1}{2} \times 1 \right) ^{2} - \left( \frac{1}{2} \times 1 \right) ^{2}+1 \\ \small y= x^2 + x+ \left( \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ \small y= \left( x+ \frac{1}{2} \right) ^{2} - \left( \frac{1}{2} \right) ^{2}+1 \\ \small y= \left( x+ \frac{1}{2} \right) ^{2} + \frac{3}{4} \end{gather*}The perfect-square trinomial was \(x^2+x+(\frac{1}{2})^2\)
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\(a=3\), so we need to factor out by 3.
\begin{multline*} \small y=3x^2+7x-6\\ \small y=3 \left( x^2 + \frac{7x}{3} - 2 \right) \\ \small y= 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] \\ \small y= 3 \left[ x^2 + \frac{7x}{3}+ \left( \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ \small y= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \left( \frac{7}{6} \right) ^{2}-2 \right] \\ \small y= 3 \left[ \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{36} \right]\\ \small y= 3 \left( x+ \frac{7}{6} \right) ^{2} - \frac{121}{12} \end{multline*}
Solutions
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 in one step
- 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.
\begin{gather*} \small i. x^2+x+1=0\\ \small ii. 3x^2+7x-6=0 \end{gather*}\(x^2+x+1=0\)
\begin{gather*} \small x^2 +x+ \left( \frac{1}{2} \times 1 \right) ^{2} = \left( \frac{1}{2} \times 1 \right) ^{2}-1 \\ \small x^2 + x+ \left( \frac{1}{2} \right) ^{2} = \left( \frac{1}{2} \right) ^{2}-1 \\ \small \left( x+ \frac{1}{2} \right) ^{2} = - \frac{3}{4}\\ \small \sqrt{\left( x+ \frac{1}{2} \right) ^{2}} = \pm \sqrt{ \frac{-3}{4}}\\ \small x+ \frac{1}{2} = \pm \frac{{\sqrt{-3}}}{4}\\ \small x=- \frac{1}{2} \pm \frac{{\sqrt{-3}}}{4}\\ \small x = \frac{{-1 \pm \sqrt{-3}}}{4} \end{gather*}The solutions to this expression are non-real. This is beyond our scope.
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\(3x^2+7x-6=0\)
\begin{gather*} \small 3x^2+7x=6\\ \small x^2+\frac{7}{3}{x}=2\\ \small 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\\ \small x^2 + \frac{7x}{3}+ \left( \frac{7}{6} \right) ^{2} = \left( \frac{7}{6} \right) ^{2}+2 \\ \small \left( x+ \frac{7}{6} \right) ^{2} = \frac{121}{36} \\ \small \sqrt{\left( x+ \frac{7}{6} \right) ^{2}} = \pm \sqrt{\frac{121}{36}}\\ \small x+ \frac{7}{6} = \pm \frac{11}{6} \\ \small x=- \frac{7}{6} \pm \frac{11}{6} \\ \small x= \frac{2}{3} \textrm{ or } x = -3 \end{gather*}Once you get used to this procedure, the steps can be shortened. Ours are long because we are trying to explain to you
Solutions