A Level Algebra and Functions, A Level Maths, KS5

Completing the Square – AS Level Maths

‘Completing the square’ is a method of solving quadratic equations by re-writing a quadratic expression as a squared expression and a constant – the quadratic formula is based on completing the square for the equation ax^2 + bx + c = 0 . It also allows you to find key information about quadratic graphs, the turning point and roots, needed to sketch them.

The examples below explain how to complete the square and how to use completing the square to sketch a quadratic graph. Click here for a completing the square resource pack, including worked examples, questions and answers in PDF and slideshow formats. To practise some of the required knowledge for completing the square, click here, or click here for a selection of completing the square activity mats.

Example Question 1
Write x^2 + 6x + 4 in the form (x+a)^2 + b where a and b are integers.

First, find the value of a which will result in 6x when we expand (x + a)^2 – we can do this by halving the coefficient of x , which gives us 3:

\begin{aligned} (x + 3)^2 &= (x + 3)(x + 3) \\ &= x^2 + 6x + 9 \end{aligned}

The first two terms are correct, but we want ” + 4″ rather than “+ 9”. To correct this, simply subtract 5, to give:

\boldsymbol{x^2 + 6x + 4 = (x + 3)^2 - 5}

This expands to:

\begin{aligned} (x+3)^2 - 5 &= (x+3)(x+3) - 5 \\ &= x^2 + 6x + 9 - 5 \\ &= x^2 + 6x + 4 \end{aligned}

More generally, the expansion of (x \pm a)^2 will always give a constant term of +a^2. Therefore, when we complete the square to factorise the x^2 and x terms, we can subtract a^2 from the constant term to give:

\boldsymbol{(x + 3)^2 - 3^2 + 4 = (x + 3)^2 - 5}

Example Question 2
By completing the square, find the roots of the equation y = 3x^2 + 12x + 1

In this example, unlike example 1, the coefficient of x^2 isn’t 1. To be able to complete the square, we need to remove this as a factor:

y = 3x^2 + 12x + 1 = y = 3[x^2 + 4x] + 1

Now, we complete the square inside the brackets as usual. Start by halving the coefficient of x and subtracting its square:

\begin{aligned} y &= 3[x^2 + 4x] + 1 \\ &= 3[(x + 2)^2 - 2^2] + 1 \end{aligned}

Once we have completed the square inside the square brackets, we can expand and simplify our expression:

\begin{aligned} y &= 3x^2 + 12x + 1 \\ &= 3[(x+2)^2 - 4] + 1 \\ &= 3(x+2)^2 - 12 + 1 \\ &= 3(x+2)^2 - 11 \end{aligned}

Finally, we can find the 2 solutions. The solution of a quadratic equation is where y = 0 , so we set y as 0 and solve:

3(x+2)^2 - 11 = 0
3(x+2)^2 = 11
(x+2)^2 = \frac{11}{3}
x + 2 = \pm \sqrt{\frac{11}{3}}
\boldsymbol{x = -2 \pm \sqrt{\frac{11}{3}}}

Of course, it can be helpful to have the roots in exact form. If needed, we can write out solutions as decimals:

x = -0.085 and x = -3.91 (to 3 s.f.).

Example Question 3
By completing the square, find the turning point and the equation of the line of symmetry of y = x^2 +5x - 3 .

As before, start by halving the coefficient of x and adjusting our constant term:

\begin{aligned} y &= x^2 + 5x - 3 \\ &= (x + \frac{5}{2})^2 - (\frac{5}{2})^2 - 3 \\ &= (x + \frac{5}{2})^2 - \frac{25}{4} - 3 \\ &= (x + \frac{5}{2})^2 - \frac{37}{4} \end{aligned}

Now we’ve completed the square, we need to find the turning point of the graph.

A quadratic graph is either a U shape (if the coefficient of x^2 is positive) or a ∩ shape (if the coefficient is negative).

As this equation has a positive coefficient of x^2 , the turning point is the coordinate of the bottom of the U where the gradient changes from negative to positive. It is also a minimum – the lowest y -value the function can have.

Consider the smallest value y = (x + \frac{5}{2})^2 - \frac{37}{4} can have.

Since a square cannot be negative, the lowest value (x + \frac{5}{2})^2 can have is 0. We can find the turning point by evaluating x and y when this is the case:

\begin{aligned} y &= 0 - \frac{37}{4} \\ &= - \frac{37}{4} \end{aligned}

This gives us our y -value. We have already set (x + \frac{5}{2})^2 equal to 0, so:

(x + \frac{5}{2})^2 = 0 \\ x + \frac{5}{2} = 0 \\ x = -\frac{5}{2}

The mimimum point of y = x^2 + 5x - 3 is at (-\frac{5}{2}, -\frac{37}{4}) .

The line of symmetry is the vertical line through the turning point. In this case, the line of symmetry will be the line with equation x = -\frac{5}{2} .

Example Question 4
By completing the square, determine the turning point of the graph y = -x^2 + 4x - 5 and therefore the number of roots.

It is important to notice that x^2 has a coefficient of -1. As before, this must be removed as a factor:

\begin{aligned} -x^2 + 4x - 5 &= -[x^2 - 4x] - 5 \\ &= -[(x - 2)^2 - 4] - 5 \\ &= -(x - 2)^2 + 4 - 5 \\ &= -(x - 2)^2 - 1 \end{aligned}

It is now possible to determine the turning point. Since the coefficient of x^2 is negative, the equation is a negative quadratic. This means the turning point will be a maximum point.

The maximum point is where -(x - 2)^2 = 0 , so x = 2 . When this is the case, y = -1 so our turning point is at (2, -1).

The maximum point is below the x-axis. This means the graph will never be above the x-axis and will have no roots. We can easily show that this is the case by considering when y = 0:

-x^2 +4x - 5 =0 \\ -(x-2)^2 - 1 = 0 \\ (x - 2)^2 = -1 \\ x - 2 = \pm \sqrt{-1}

This has no real solutions, therefore -x^2 +4x - 5 = 0 has no real roots.

We have looked at how to complete the square and how to use the result to find the turning point, line of symmetry, and roots of a quadratic equation. Now, let’s look at how to sketch a graph with all this information.

Example Question 5
Sketch the graph of y = 2x^2 - 7x + 5 , giving the coordinates of the turning point of the graph and any points where the graph crosses the axes.

Start by completing the square:

\begin{aligned} 2x^2 - 7x + 5 &= 2[x^2 - \frac{7}{2}x] + 5 \\ &= 2[(x - \frac{7}{4})^2 - \frac{49}{16}] + 5 \\ &= 2(x - \frac{7}{4})^2 - \frac{49}{8} + 5 \\ &= 2(x - \frac{7}{4})^2 - \frac{9}{8} \end{aligned}

This shows that the turning point of the graph is at (\frac{7}{4}, -\frac{9}{8}) and, since the graph is a positive quadratic, this is a minimum point.

Next, find the roots of the equation:

2(x - \frac{7}{4})^2 - \frac{9}{8} = 0 \\ 2(x - \frac{7}{4})^2 = \frac{9}{8} \\ (x - \frac{7}{4})^2 = \frac{9}{16} \\ x - \frac{7}{4} = \pm \frac{3}{4} \\ x = \frac{7}{4} \pm \frac{3}{4}

Therefore, the two solutions are x = \frac{5}{2} and x = 1 .

Finally, we need to find any points where the graph crosses the axes; this includes the y-axis.

The y-intercept can be read straight from the original equation, y = 2x^2 - 7x + 5 . When x = 0, y = 5 , so the y-intercept is (0, 5).

We now have enough information to draw a sketch. Since it’s a quadratic, the graph will be a parabola and we know the positions of the y- and x-intercepts and the turning point of the graph.

The Quadratic Formula

The quadratic formula can be derived by using completing the square to rearrange the quadratic equation ax^2 + bx + c = 0 into the form x = ... .

Try this yourself, or check below for the working.

First, complete the square:

ax^2 + bx + c = 0 \\ a[x^2 + \frac{bx}{a}] + c = 0 \\ a[(x + \frac{b}{2a})^2 - \frac{b^2}{4a^2}] + c = 0 \\ a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c = 0 \\ a(x + \frac{b}{2a})^2 - \frac{b^2\;-\;4ac}{4a} = 0

Then, rearrange:

a(x + \frac{b}{2a})^2 = \frac{b^2\;-\;4ac}{4a} \\(x+\frac{b}{2a})^2 = \frac{b^2\;-\;4ac}{4a^2} \\x+\frac{b}{2a} = \pm \sqrt{\frac{b^2\;-\;4ac}{4a^2}} \\ x+\frac{b}{2a} = \frac{\pm \sqrt{b^2\;-\;4ac}}{2a} \\ x = - \frac{b}{2a} + \frac{\pm \sqrt{b^2\;-\;4ac}}{2a} \\ x = \frac{-b\ \pm \sqrt{b^2\;-\;4ac}}{2a}

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