Solving quadratics

Tier: #Foundation #Higher

🔗What you need to know first
How to

A quadratic equation has the form $ax^2 + bx + c = 0$. There are three main methods for solving them.

Method 1 — Factorising (fastest when it works): $$x^2 + 5x + 6 = 0 \Rightarrow (x+2)(x+3) = 0 \Rightarrow x = -2 \text{ or } x = -3$$

Method 2 — Quadratic formula (always works): $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Example: $2x^2 - 3x - 5 = 0$ $$x = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm 7}{4} \Rightarrow x = 2.5 \text{ or } x = -1$$

Method 3 — Completing the square (useful for exact answers or the vertex): $$x^2 + 6x + 1 = 0 \Rightarrow (x+3)^2 - 8 = 0 \Rightarrow x = -3 \pm 2\sqrt{2}$$

The discriminant $b^2 - 4ac$ tells you how many solutions:

  • $> 0$: two solutions
  • $= 0$: one repeated solution
  • $< 0$: no real solutions

Common error: forgetting the $\pm$ in the formula, giving only one solution.

Questions to practise

Practise these questions →

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📝Past paper questions
💬What the examiners say
  • "The approach of eliminating y to get an equation in x is more straightforward than eliminating x."
  • "Making y the subject was the sensible approach; sign errors in substitution and rearrangement cost marks; completing the working by finding matching y values and writing pairs clearly was necessary for full marks."
⬆️How you can quickly improve
  • Before factorising or using the formula, rearrange so all terms are on one side and the equation equals zero.
  • Expand squared brackets term by term — (a − b)² = a² − 2ab + b² — write all three terms explicitly, then check by expanding back.
  • Always give both solutions. After factorising, set each bracket equal to zero and solve them separately.
  • Read the question before picking a method: if area is involved, use the area formula; if a missing side, use Pythagoras. Write the formula before substituting.
🔓What this unlocks
ℹ️Calculator tricks

Store $b^2 - 4ac$ in memory first to avoid errors in the formula.