# Solving by completing the square - Higher

Some quadratics cannot be factorised. An alternative method to solve a quadratic equation is to complete the square.

To solve an equation of the form $$x^2 + bx + c = 0$$ consider the expression $$\left(x + \frac{b}{2}\right)^2 + c$$.

$$\left(x + \frac{b}{2}\right)^2 + c$$ expands to $$x^2 + bx + \left(\frac{b}{2}\right)^2 + c$$, which is the same as the left-hand side of the original equation but with an extra term $$\left(\frac{b}{2}\right)^2$$.

To get back to the correct original expression, this extra term has to be subtracted.

So the equation $$x^2 + bx + c = 0$$ becomes $$\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c = 0$$.

This can be rearranged to give $$\left(x + \frac{b}{2}\right)^2 = \left(\frac{b}{2}\right)^2 - c$$ which can then be solved by taking the square root of both sides.

### Example

Solve $$x^2 + 6x - 10 = 0$$ by completing the square.

Start by following the formula: $$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$$

For this example, this gives: $$\left(x + \frac{6}{2}\right)^2 - \left(\frac{6}{2}\right)^2 - 10 = 0$$

This will simplify to: $$(x + 3)^2 - (3)^2 - 10 = 0$$

This simplifies further to $$(x + 3)^2 - 9 - 10 = 0$$, which simplifies further to $$(x + 3)^2 - 19 = 0$$.

Rearrange this quadratic to get $$(x + 3)^2$$ alone on the left hand side by adding 19 to each side.

This is the solution to the question in form, which gives the exact values of the solutions. If you are asked for exact solutions, leave your answer in surd form.

To find approximate solutions in decimal form, continue on with a calculator, adding and subtracting the square root to find the two solutions.

$x = -3 \pm \sqrt{19}$

The first solution is: $$x = -3 + \sqrt{19} = 1.36$$ (3 sf)

The second solution is: $$x = -3 - \sqrt{19} = -7.36$$ (3 sf)

Question

Solve $$x^2 - 4x - 3 = 0$$ by completing the square.

To complete the squares in the form of $$ax^2 + bx + c = 0$$: use $$\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$$

$x^2 - 4x - 3 = 0 \rightarrow \left(x - \frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2 - 3 = 0$

$\rightarrow (x - 2)^2 - 2^2 - 3 = 0 \rightarrow (x - 2)^2 - 4 - 3 = 0 \rightarrow (x - 2)^2 - 7 = 0$

$\begin{array}{rcl} (x - 2)^2 - 7 & = & 0 \\ +7 && +7 \\ (x - 2)^2 & = & 7 \\ \sqrt{} && \sqrt{} \\ x - 2 & = & \pm \sqrt{7} \\ +2 && +2 \\ x & = & 2 \pm \sqrt{7} \end{array}$

This gives the two solutions in surd form. To find approximate solutions in decimal form, continue to work the answer out with a calculator.

$x = 2 \pm \sqrt{7}$

$$x = 2 + \sqrt{7}$$ or $$x = 2 - \sqrt{7}$$

$$x = 4.65$$ (3 sf) or $$x = -0.65$$ (3 sf)