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

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

When there’s a number in front of x²

Take it out as a factor of the $x$ terms, or divide the whole equation by it. Don’t divide an expression that isn’t set equal to 0: that changes it.

For $3y^2 - 5y + 2 = 0$, divide through by 3:

so $y - \frac56 = \pm\frac16$, giving $y = 1$ or $y = \frac23$.

Using the perfect squares

The same picture gives two expansions worth knowing by heart: $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$. An expression is a perfect square when it fits one of these exactly, and they link $a^2 + b^2$ with $a + b$, $a - b$ and $ab$. For example, if $a + b = 5$ and $ab = 4$, then $a^2 + b^2 = (a + b)^2 - 2ab = 25 - 8 = 17$.

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 $(x + p)^2$ is smallest (0) when $x = -p$. Then $y = q$.

A past question, step by step

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