Completing the square

When factorising is not enough

Try to factorise $x^2 - 4x + 1$. We need two numbers whose product is $1$ and whose sum is $-4$. No whole numbers work, and no simple fractions either. Yet the equation $x^2 - 4x + 1 = 0$ does have roots. We need a method that works every time.

The idea comes from equations like

This one is easy. A number whose square is $3$ is $\sqrt{3}$ or $-\sqrt{3}$, so $x - 2 = \pm\sqrt{3}$ and $x = 2 \pm \sqrt{3}$. If we can turn any quadratic equation into the form $(x + p)^2 = k$, we can solve it. Turning it into that form is called completing the square.

A picture of the method

Draw $x^2 + 6x$ as an area. The $x^2$ is a square of side $x$. Split the $6x$ into two strips, each $3$ by $x$, and put one along each side of the square.

The shape is almost a square of side $x + 3$. Only the corner is missing, a small square of side $3$ and area $9$. Add $9$ and the square is complete:

The missing piece is always the square of half the coefficient of $x$.

The method, step by step

Solve $x^2 + 6x - 7 = 0$.

1. Move the number term. $x^2 + 6x = 7$.
2. Add the square of half of $b$ to both sides. Half of $6$ is $3$, and $3^2 = 9$. So $x^2 + 6x + 9 = 16$.
3. Write the left side as a square. $(x + 3)^2 = 16$.
4. Take the square root, with both signs. $x + 3 = 4$ or $x + 3 = -4$.
5. Solve. $x = 1$ or $x = -7$.

Check. $1^2 + 6 \times 1 - 7 = 0$, and $(-7)^2 + 6 \times (-7) - 7 = 49 - 42 - 7 = 0$.

Now the equation that would not factorise: $x^2 - 4x + 1 = 0$. Half of $-4$ is $-2$, and $(-2)^2 = 4$.

So $x \approx 3.732$ or $x \approx 0.268$.

When the square term has a number in front

Solve $2x^2 - 8x + 3 = 0$. First divide every term by $2$, so that $x^2$ stands alone:

Then complete the square as before: $(x - 2)^2 = 4 - \tfrac{3}{2} = \tfrac{5}{2}$. So $x = 2 \pm \sqrt{\tfrac{5}{2}}$, which is about $3.58$ or $0.42$.

When there is no answer

Try $x^2 + 2x + 5 = 0$. Completing the square gives $(x + 1)^2 = -4$. But the square of a real number is never negative. So this equation has no real roots. The method tells us so, without any guessing.

Practise with new numbers

Try it in Python

The program completes the square for three equations and prints the roots, or says that there are none.

Common mistakes

• Adding $b^2$ instead of $\left(\frac{b}{2}\right)^2$. For $x^2 + 6x$ we add $9$, not $36$.
• Adding to one side only. Whatever is added to the left must be added to the right too.
• Forgetting the negative root. $(x + 3)^2 = 16$ gives $x + 3 = 4$ or $x + 3 = -4$.
• Not dividing first. With $2x^2$, divide every term by $2$ before you halve the coefficient of $x$.

Remember these

• $x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$.
• Move the number term, add $\left(\frac{b}{2}\right)^2$ to both sides, take the square root with $\pm$, and solve.
• If the right side comes out negative, there are no real roots.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which number completes the square for $x^2 - 12x$?

2. Solve $(x - 4)^2 = 9$.

3. Write $x^2 + 2x + 7$ in the form $(x + p)^2 + q$.

4. Why does $x^2 + 4x + 9 = 0$ have no real roots?

5. To solve $3x^2 + 6x - 2 = 0$ by completing the square, what is the first step?