Solve Quadratic Equation by Completing the Square

A quadratic equation can appear in many different formats. A quadratic equation is one that can be rearranged to the form ax^2+bx+c=0.

Solving this equation is the same process as finding the x intercepts of the function y=ax^2+bx+c.

The shape of the graph of a quadratic function is a parabola.

A parabola may cross the x-axis twice:

The figure above is an example of two, real, distinct solutions to the equation ax^2+bx+c=0.

A parabola may just touch the x-axis, with the x-axis being a tangent to the turning point of the graph:

The figure above is an example of there being one, real repeated solution to the equation ax^2+bx+c=0.

A parabola may not cross the x-axis at all:

The figure above is an example of there being no real solutions to the equation ax^2+bx+c=0.

In the context of graphing, solving a quadratic equation leads to the roots (x-intercepts) of the parabola.

Solving

When all three terms of the quadratic expression are present, we need to use factoring, the quadratic formula or the completing square method to solve.

We can solve linear equations more directly – we can generally isolate the x just by a sequence of adding, subtracting, multiplying, dividing both sides by the same quantity. However, in a quadratic equation we have both an x term and an x^2 term which makes isolating x more difficult.

When there are only two terms, or if the expression factors

If a quadratic equation has only the x^2 term and the constant, the equation is not difficult.

Example (a)

    \begin{align*}x^2-8&=0\\[10 pt]x^2&=8\\[10 pt]x&=\pm\sqrt{8}\end{align}

Also, if a quadratic equation  has only the x^2 term and the x term, it can be factored.

Example (b)

    \begin{align*}x^2-8x&=0\\[10 pt]x(x-8)&=0\\[10 pt]x=0, \quad x&=8\end{align}

The third kind of quadratic equation that is not difficult is the kind that factors easily.

Example (c)

    \begin{align*}x^2-15x+36&=0\\[10 pt](x-3)(x-12)&=0\\[10 pt]x=3,\quad x&=12\end{align}

In these three examples, there is no need for either the quadratic formula or the completed square method.

Completing the Square

If there is no fast way to the solution then we must use either the quadratic formula or the completed square method.

Example 1, expression only

    \[x^2-8x+5\]

The process is to introduce a square term (using half of -8, the coefficient of the x term) and then to subtract any new terms introduced by this process:

    \[(x-4)^2+5=(x-4)(x-4)+5=x^2-8x\textbf{+16}+5\]

We see that using (x-4)^2 instead of x^2-8x we have introduced an extra +16. So we need to subtract the 16 in order for our final line to equal our original line.

    \[x^2-8x+5=(x-4)^2-16+5=(x-4)^2-11\]

Now let’s back up to make sure the right hand side is equal to the left hand side:

    \[(x-4)(x-4)-11=x^2-8x+16-11=x^2-8x+5\,\,\checkmark\]

Therefore

    \[x^2-8x+5=(x-4)^2-11\]

This is called the completed square form. It is useful to us in the context of solving a quadratic equation because the unknown x appears only once, which makes it possible to isolate.

Using Completed Square to Solve

Example 1

Now lets consider the equation x^2-8x+5=0. By completing the square we can solve as follows:

    \begin{align*}x^2-8x+5&=0\\[10 pt](x-4)^2-16+5&=0\\[10 pt](x-4)^2-11&=0\\[10 pt](x-4)^2&=11\\[10 pt]x-4&=\pm\sqrt{11}\\[10 pt]x&=4\pm\sqrt{11}\end{align}

Example 2

    \begin{align*}x^2+10x-3&=0\\[10 pt](x+5)^2-25-3&=0\\[10 pt](x+5)^2-28&=0\\[10 pt](x+5)^2&=28\\[10 pt]x+5&=\pm\sqrt{28}\\[10 pt]x&=-5\pm\sqrt{4\times 7}\\[10 pt]x&=-5 \pm 2\sqrt{7}\end{align}

Example 3

    \begin{align*}x^2+7x-1&=0\\[10 pt]\left(x+\frac{7}{2}\right)^2-\frac{49}{4}-1&=0\\[10 pt]\left(x+\frac{7}{2}\right)^2-\frac{49}{4}-\frac{4}{4}&=0\\[10 pt]\left(x+\frac{7}{2}\right)^2-\frac{53}{4}&=0\\[10 pt]\left(x+\frac{7}{2}\right)^2&=\frac{53}{4}\\[10 pt]x+\frac{7}{2}&=\pm\sqrt{\frac{53}{4}}\\[10 pt]x&=-\frac{7}{2}\pm\frac{\sqrt{53}}{2}\end{align}

More Difficult: When the coefficient of the x^2 term is not 1

Example 4

Solve 3x^2-4x-5=0

Because we have an equation, it is permitted to divide all terms by 3 as follows:

    \begin{align*}3x^2-4x-5&=0\\[10 pt] x^2-\frac{4}{3}x-\frac{5}{3}&=\frac{0}{3}\\[10 pt]x^2-\frac{4}{3}x-\frac{5}{3}&=0\end{align}

We continue as follows:

    \begin{align*}x^2-\frac{4}{3}x-\frac{5}{3}&=0\\[10 pt]\left(x-\frac{2}{3}\right)^2-\frac{4}{9}-\frac{5}{3}&=0\\[10 pt]\left(x-\frac{2}{3}\right)^2-\frac{4}{9}-\frac{15}{9}&=0\\[10 pt]\left(x-\frac{2}{3}\right)^2-\frac{19}{9}&=0\\[10 pt]\left(x-\frac{2}{3}\right)^2&=\frac{19}{9}\\[10 pt]x-\frac{2}{3}&=\pm\sqrt{\frac{19}{9}}\\[10 pt]x&=\frac{2}{3} \pm \frac{\sqrt{19}}{3}\end{align}

Example 5

We can follow the same process if the coefficient of x^2 is negative:

    \begin{align*}-x^2-9x+5&=0\\[10 pt] x^2+9x-5&=0\\[10 pt]\left(x+\frac{9}{2}\right)^2-\frac{81}{4}-5&=0\\[10 pt]\left(x+\frac{9}{2}\right)^2-\frac{81}{4}-\frac{20}{4}&=0\\[10 pt]\left(x+\frac{9}{2}\right)^2-\frac{101}{4}&=0\\[10 pt]\left(x+\frac{9}{2}\right)^2&=\frac{101}{4}\\[10 pt]x+\frac{9}{2}&=\pm\sqrt{\frac{101}{4}}\\[10 pt]x&=-\frac{9}{2} \pm \frac{\sqrt{101}}{2}\end{align}

Example 6

Solve x^2+6x+10=0

    \begin{align*}(x+3)^2-9+10&=0\\[10 pt](x+3)^2+1&=0\\[10 pt](x+3)^2&=-1\\[10 pt]x+3&=\pm\sqrt{-1}\quad\dots \text{error}\end{align}

As there are no real solutions to \sqrt{-1}, we can conclude that there are no x-intercepts to the parabola y=x^2+6x+10, and therefore, there are no real solutions to the equation x^2+6x+10=0.

Practice

Practice using the completed square method to find the roots of these equations. Enter your answers correct to two decimal places.

applet link

Top BC G11 Menu Number and Algebra Menu