Solving Quadratic EquationsEdexcel IGCSE Maths: Revision notes

A quadratic equation has the general form $ax^2+bx+c=0$ where $a\neq 0$. The highest power of $x$ is 2, so there are usually two solutions (roots), though sometimes one repeated root or none (over the reals).

Before solving, always rearrange the equation so that everything is on one side and it equals zero: $3x^2+5x=2 \implies 3x^2+5x-2=0$

Factorisation works when $ax^2+bx+c$ splits into two brackets. For $a=1$, find two numbers that multiply to give $c$ and add to give $b$.

Example: solve $x^2+5x+6=0$.

• Numbers that multiply to 6 and add to 5: 2 and 3.
• So $(x+2)(x+3)=0$.
• Using the Zero Product Rule: $x+2=0$ or $x+3=0$, so $x=-2$ or $x=-3$.

When $a\neq 1$ (e.g. $2x^2+7x+3=0$), use the same idea but split the middle term: find two numbers multiplying to $a \times c = 6$ and adding to $b=7$ (these are 1 and 6). Rewrite: $2x^2+x+6x+3=0$, factorise in pairs: $x(2x+1)+3(2x+1)=0$, giving $(2x+1)(x+3)=0$.

Use the quadratic formula, which works for every quadratic: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

First identify $a$, $b$ and $c$ from $ax^2+bx+c=0$, then substitute carefully — brackets around negative values help avoid sign errors.

Example: solve $2x^2+3x-4=0$ ($a=2$, $b=3$, $c=-4$). $x=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)}=\frac{-3\pm\sqrt{9+32}}{4}=\frac{-3\pm\sqrt{41}}{4}$ Give answers to an appropriate degree of accuracy (e.g. 2 or 3 significant figures) unless told to leave in surd form.

The discriminant $b^2-4ac$ predicts the number of real roots without fully solving:

• If $b^2-4ac>0$: two distinct real roots.
• If $b^2-4ac=0$: one repeated root (the curve touches the x-axis).
• If $b^2-4ac<0$: no real roots (the curve doesn't cross the x-axis).

This links directly to the shape of the quadratic graph, so exam questions often ask you to use the discriminant to determine how many times a curve meets the x-axis, or to find a value of a constant for which an equation has equal roots.

Completing the square rewrites $ax^2+bx+c$ in the form $a(x+p)^2+q$. For $a=1$: halve $b$ to get $p$, then $x^2+bx+c=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c$

Example: $x^2+6x+2=0$ $(x+3)^2-9+2=0 \implies (x+3)^2=7 \implies x=-3\pm\sqrt{7}$

This method is also used to find the turning point of a quadratic graph: for $y=(x+p)^2+q$, the minimum (or maximum) is at $(-p, q)$.

When $a\neq 1$, factor $a$ out of the $x^2$ and $x$ terms first before completing the square inside the bracket.