Quadratic equations

Solve quadratics by factorising, the quadratic formula and completing the square; use the discriminant and vertex form.

Also known as: factorising quadratics, quadratic formula, completing the square, parabolas.

SAT · Advanced Math GCSE Maths · Algebra IGCSE Maths · Algebra and graphs

Key ideas

Rearrange to $ax^2 + bx + c = 0$ before solving.
If a product is zero, one of the factors must be zero.
Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Discriminant $b^2 - 4ac$ : positive → 2 roots, zero → 1 root, negative → no real roots.
Vertex form $y = a(x - h)^2 + k$ has turning point $(h, k)$ .

Solving by factorising

Write the equation as $ax^2 + bx + c = 0$ . For $x^2 - 5x + 6 = 0$ , find two numbers that multiply to $6$ and add to $-5$ : $-2$ and $-3$ .

$(x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3$

When $a \neq 1$ , look for factors like $(2x - 1)(x - 3)$ and check by expanding.

The quadratic formula and discriminant

When factorising is hard, use $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$ For $x^2 - 4x - 1 = 0$ : $x = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$ .

The discriminant $b^2 - 4ac$ tells you how many real solutions there are, without solving. Also useful: the solutions sum to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$ .

Completing the square and vertex form

$x^2 + 6x + 5 = (x + 3)^2 - 4$ . In vertex form $y = a(x - h)^2 + k$ , the graph's turning point is $(h, k)$ , so $y = (x + 3)^2 - 4$ has its minimum at $(-3, -4)$ .

Worked examples

Example 1 · Factorising with a ≠ 1

Solve $2x^2 - 7x + 3 = 0$ .

Step 1.Look for $(2x + p)(x + q)$ with $pq = 3$ and $2q + p = -7$ : $(2x - 1)(x - 3)$ .
Step 2. $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ ; $x - 3 = 0 \Rightarrow x = 3$ .

Answer: $x = \frac{1}{2}$ or $x = 3$

Example 2 · Using the discriminant

For what value of $c$ does $x^2 + 6x + c = 0$ have exactly one real solution?

Step 1.One solution when $b^2 - 4ac = 0$ : $36 - 4c = 0$ .
Step 2. $c = 9$ . Check: $x^2 + 6x + 9 = (x + 3)^2$ .

Answer: $c = 9$

Common mistakes

Trying to solve before the equation equals zero.
Sign errors in the formula: it is $-b$ , and $b^2$ is always non-negative.
Dividing both sides by $x$ and losing the solution $x = 0$ .

Try it yourself

Original practice questions. Your answer is checked instantly.

Question 1Quadratic equationsEasier

What are the solutions of $x^2 - 5x + 6 = 0$ ?

Question 2Quadratic equationsMedium

What is the positive solution of $x^2 + 2x - 15 = 0$ ?