Quadratic EquationsCambridge IGCSE Maths: Revision notes

A quadratic equation has the form $ax^2 + bx + c = 0$. If it factorises, use the fact that if two factors multiply to give zero, at least one of them must be zero.

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

1. Factorise: $(x+2)(x+3) = 0$
2. Set each factor to zero: $x+2=0$ or $x+3=0$
3. Solve: $x = -2$ or $x = -3$

Always rearrange the equation so one side equals zero before factorising.

Completing the square rewrites $x^2 + bx + c = 0$ as $(x+p)^2 + q = 0$, which can then be solved directly.

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

1. Complete the square: $(x+3)^2 - 9 + 5 = 0 \Rightarrow (x+3)^2 = 4$
2. Square root both sides: $x + 3 = \pm 2$
3. Solve: $x = -1$ or $x = -5$

When a quadratic doesn't factorise easily, use the quadratic formula (given in the exam), written as x equals negative b plus or minus the square root of (b squared minus 4ac), all over 2a:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $ax^2 + bx + c = 0$, identify $a$, $b$ and $c$ carefully (including signs), then substitute.

Example: solve $2x^2 + 3x - 5 = 0$ ($a=2, b=3, c=-5$): $x = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm 7}{4}$

Giving $x = 1$ or $x = -2.5$. If $b^2-4ac$ is not a perfect square, the answer should be left in surd form.

When one equation is linear and the other is quadratic (or otherwise non-linear), use substitution: rearrange the linear equation to make one variable the subject, then substitute into the quadratic equation.

Example: solve $y = x + 1$ and $y = x^2 - 5$

1. Substitute: $x + 1 = x^2 - 5$
2. Rearrange to zero: $x^2 - x - 6 = 0$
3. Factorise: $(x-3)(x+2) = 0 \Rightarrow x = 3$ or $x = -2$
4. Find corresponding $y$: $x=3 \Rightarrow y=4$; $x=-2 \Rightarrow y=-1$