Completing the square

Not every quadratic factorises neatly. $x^2 + 6x + 2$ has no pair of whole numbers that multiply to 2 and add to 6. Completing the square works every time, and the picture below shows why it’s called that.

See it as area

Step through the stages, then change the number in front of $x$ and step through again. The corner is always half the number, squared.

Solving an equation this way

1. Get the $x^2$ and $x$ terms on one side and the number on the other.
2. Add $\left(\frac b2\right)^2$ to both sides, so the left side becomes a perfect square.
3. Take the square root of both sides, remembering $\pm$.
4. Solve for $x$.

For $x^2 + 6x + 2 = 0$:

• Take 2 from both sides: ${x^2 + 6x = -2}$.
• Half of 6 is 3, so add ${3^2 = 9}$ to both sides: ${x^2 + 6x + 9 = 7}$.
• The left side is now a square: ${(x + 3)^2 = 7}$.
• Square root both sides, with $\pm$: ${x + 3 = \pm\sqrt7}$.
• Take 3 from both sides: ${x = -3 \pm \sqrt 7}$.

So $x \approx -0.35$ or $x \approx -5.65$ (to 2 decimal places).

When there’s a number in front of x²

Divide the whole equation by that number first, so the $x^2$ stands alone. You can only do this with an equation (something $= 0$). With an expression on its own, take the number out as a factor of the $x$ terms instead: dividing would change its value.

For $3y^2 - 5y + 2 = 0$:

• Divide by 3: ${y^2 - \frac53 y + \frac23 = 0}$.
• Take $\frac23$ from both sides: ${y^2 - \frac53 y = -\frac23}$.
• Half of $\frac53$ is $\frac56$, so add ${\left(\frac56\right)^2 = \frac{25}{36}}$ to both sides: ${\left(y - \frac56\right)^2 = -\frac{24}{36} + \frac{25}{36} = \frac{1}{36}}$.
• Square root, with $\pm$: ${y - \frac56 = \pm\frac16}$.

So $y = \frac56 + \frac16 = 1$ or $y = \frac56 - \frac16 = \frac23$.

Using the perfect squares

The same picture gives two expansions worth knowing by heart:

An expression is a perfect square when it fits one of these exactly. They also connect $a^2 + b^2$ to $a + b$ and $ab$. For example, if $a + b = 5$ and $ab = 4$:

Finding the vertex, and the least or greatest value

Once $y$ is written as $(x + p)^2 + q$, the vertex is easy to see. A square is never negative, so the smallest $(x + p)^2$ can be is 0, when $x = -p$. At that point $y = q$.

A past question, step by step

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