The identity $(x+p)^2 = x^2+2px+p^2$ is the engine. To write $x^2+bx+c$ as $(x+p)^2+q$: halve $b$ to get $p$, then compensate by subtracting $p^2$ (since $(x+p)^2$ introduces an extra $p^2$ that wasn't in the original).

Worked example: Rewrite $x^2+10x+9$. Halve 10 → $p=5$. Compensate: $x^2+10x \equiv (x+5)^2-25$. Re-add the original $+9$: $(x+5)^2-25+9 = (x+5)^2-16$.

You can't divide an expression by $a$ — only an equation. Instead, factorise $a$ out of just the first two terms, leaving $c$ outside:

Worked example: Rewrite $y=4x^2+16x+5$.

Now distribute the 4 across both terms inside the bracket — this is the step almost everyone gets wrong:

When $b$ is odd, $p=\frac{b}{2}$ becomes a fraction — square it carefully using $(\frac{a}{b})^2=\frac{a^2}{b^2}$, e.g. for $x^2+3x+4$: $p=\frac32$, so $(x+\frac32)^2-\frac94+4 = (x+\frac32)^2+\frac74$.

For $y=a(x+p)^2+q$, since $(x+p)^2\ge0$ always, the vertex is at $(-p,q)$ — note the sign flip on $p$. If $a>0$ it's a minimum (value $q$); if $a<0$ it's a maximum.

Worked example: For $y=4(x+2)^2-11$: vertex $(-2,-11)$, a minimum since $a=4>0$.

Rearrange $(x+p)^2=k$, then take the square root of both sides, remembering $\pm$.

Worked example (exact/surd form): Solve $x^2+10x-3=0$.

Key distinction: rewriting an expression like $4x^2+16x+5$ requires factorising $a$ out (above); solving an equation like $4x^2+16x+5=0$ lets you divide the whole equation by $a$ first, which is often simpler.

Complete the square, then argue from the fact that a square is never negative.

Worked example: Show $x^2+6x+10$ is always positive.

Since $(x+3)^2\ge0$ for all real $x$, we have $(x+3)^2+1\ge1>0$. So the expression is always positive, with minimum value 1.

The same logic with a negative leading coefficient proves "always negative": completing the square on $-x^2-4x-5$ gives $-(x+2)^2-1$, whose maximum value is $-1<0$.

(a) Express $3x-2x^2-1$ in the form $a(x+p)^2+q$. (b) Hence state the turning point. (c) Solve $3x-2x^2-1=0$ as exact fractions.

(a) Reorder: $y=-2x^2+3x-1 = -2[x^2-\frac32x]-1$. Complete the square inside: $x^2-\frac32x \equiv (x-\frac34)^2-\frac{9}{16}$.

(b) Turning point: $(\frac34,\ \frac18)$ — read directly off (a), no calculation needed.

(c) $-2(x-\frac34)^2+\frac18=0 \implies (x-\frac34)^2=\frac{1}{16} \implies x-\frac34=\pm\frac14$, giving $x=1$ or $x=\frac12$.