Tag Archives: completing the square

November 11, 2025 · 10:19 am

Completing the Square

$x^2+6x-4 \rightarrow (x+3)^2-13$

Completing the square is useful to

sketch parabolas.
solve quadratics.
factorising quadratics
finding the centre and radius version of the equation of a circle.

When completing the square we take advantage of perfect squares. For example, $(x+3)^2=(x+3)(x+3)=x^2+6x+9$

$6=2\times 3$ and $9=3\times 3$

Example 1

Put $x^2+8x-5$ into completed square form.

What perfect square has an $8x$ term?

$(x+4)^2=x^2+8x+16$

We don’t want $+16$ , we want $-5$ , so subtract $16+5$

$x^2+8x-5=(x+4)^2-21$

$x^2+bx+c=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c$

What about a non-monic quadratic? For example,

$2x^2+12x+11$

Factorise the $2$

$2(x^2+6x+\frac{11}{2})$

And continue as before

$2[(x+3)^2-9+\frac{11}{2}]=2[(x+3)^2-\frac{18}{2}+\frac{11}{2}]=2[(x+3)^2-\frac{7}{2}]=2(x+3)^2-7$

Example 2

$y=2x^2+7x-5$

$2(x^2+\frac{7}{2}x-\frac{5}{2})$

$2[(x+\frac{7}{4})^2-(\frac{7}{4})^2-\frac{5}{2}]$

$2[(x+\frac{7}{4})^2-\frac{49}{16}-\frac{40}{16}]$

$2[(x+\frac{7}{4})^2-\frac{89}{16}]$

$2(x+\frac{7}{4})^2-\frac{89}{8}$

$ax^2+bx+c=a(x+\frac{b}{2a})^2-a((\frac{b}{2a})^2+\frac{c}{a})$

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Filed under Algebra, Arithmetic, Classpad Skills, Completing the Square, Fractions, Quadratic, Quadratics, Year 10 Mathematics, Year 9 Mathematics

Tagged as completing the square, quadratics

March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is $y=ax^2+bx+c$

Completing the square,

$\begin{equation*}ax^2+bx+c\end{equation}$

Factorise out the leading coefficient (i.e. $a$ )

$\begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}$

Half the second term (i.e $\frac{b}{a}$ ) and subtract the square of the second term.

$\begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}$

Simplify

$\begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}$

$\begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}$

Now let’s solve

$\begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}$

$\begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}$

$\begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

Which is the quadratic equation formula.

Filed under Algebra, Quadratic, Quadratics, Solving, Solving, Solving Equations

Tagged as completing the square, quadratic equation formula