Completing the square calculator with working

Example: $x^2 + 6x + 5$

1. Halve the coefficient of $x$: $6 \div 2 = 3$.
2. Write $(x + 3)^2$, which expands to $x^2 + 6x + 9$.
3. That's 9 too many compared with the 5 we need, so subtract: $(x + 3)^2 + 5 - 9$.
4. Answer: $(x + 3)^2 - 4$.

Turning point: $(-3, -4)$, a minimum.

Example: $2x^2 + 8x + 3$

1. Factor 2 out of the $x$ terms: $2(x^2 + 4x) + 3$.
2. Complete the square inside: $x^2 + 4x = (x + 2)^2 - 4$.
3. Multiply back: $2[(x + 2)^2 - 4] + 3 = 2(x + 2)^2 - 8 + 3$.
4. Answer: $2(x + 2)^2 - 5$, with turning point $(-2, -5)$.

From $(x + 3)^2 - 4 = 0$: $(x + 3)^2 = 4$, so $x + 3 = \pm 2$, giving $x = -1$ or $x = -5$.

When the number isn't a square, the answer is a surd: $x^2 + 4x - 1 = 0$ becomes $(x + 2)^2 = 5$, so $x = -2 \pm \sqrt{5}$. If the right-hand side is negative, there are no real solutions.

The graph of $y = a(x + p)^2 + q$ has its turning point at $(-p, q)$. If $a > 0$ it's a minimum (∪ shape); if $a < 0$ it's a maximum (∩ shape). The minimum value of $(x + 3)^2 - 4$ is $-4$, which happens when $x = -3$. Sketch it with our graphing calculator.

• Forgetting to subtract the square of the half: $(x + 3)^2$ adds an extra 9.
• Getting the sign of the turning point wrong: $(x + 3)^2$ gives $x = -3$.
• Not factoring out $a$ first when $a \neq 1$.

Practise with our completing the square questions, and compare with the quadratic equation solver.

• "Write in the form $(x + p)^2 + q$" wants $p$ and $q$ stated, so write them out.
• Use completed square form to explain why an expression is always positive: $(x + 3)^2 + 1 \geq 1$.
• Completing the square is Higher tier at GCSE and Extended at IGCSE.

The quadratic formula is completing the square done once in general. Starting from $ax^2 + bx + c = 0$, completing the square gives

So any quadratic you can solve with the formula, you can also solve by completing the square. Completing the square has the bonus of giving the turning point, which the formula doesn't.