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G5.8 Solving Quadratic Equations

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01 — Explanation

How do I solve a quadratic by factorising?

Once a quadratic is written as $(x+p)(x+q)=0$ , the Null Factor Law says at least one bracket must equal zero. Set each bracket to zero separately and solve: if $(x-3)(x+7)=0$ , then $x=3$ or $x=-7$ .

Exam Tip

If $c=0$ (e.g. $x^2-x=0$ ), do not divide both sides by $x$ — this silently deletes the solution $x=0$ . Instead factorise: $x(x-1)=0 \Rightarrow x=0$ or $x=1$ .

02 — Explanation

How do I know whether to factorise or use the quadratic formula?

Check what the question asks for: if it demands an answer to a given number of decimal places or significant figures, the roots are very unlikely to be nice integers, and attempting factorisation wastes time. Go straight to the formula:

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

If no such precision is specified and the numbers look manageable, try factorising first — it's faster and lower-risk.

03 — Explanation

How do I use the quadratic formula without making sign errors?

Always substitute negative coefficients in brackets. If $a=1, b=-8, c=-20$ :

x=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-20)}}{2(1)} = \frac{8\pm\sqrt{64+80}}{2}=\frac{8\pm12}{2}

giving $x=10$ or $x=-2$ .

Exam Tip

A very common error is dividing only the square-root term by $2a$ , rather than the entire numerator $-b\pm\sqrt{b^2-4ac}$ . Keep the whole expression as one fraction until the final step.

If a non-integer answer is required in exact surd form, simplify the surd before dividing: e.g. $\sqrt{88}=2\sqrt{22}$ , so $x=\dfrac{8\pm2\sqrt{22}}{4}$ simplifies to $x=\dfrac{4\pm\sqrt{22}}{2}$ .

04 — Explanation

What if the equation isn't already set to zero?

Many Grade 5+ questions hide the quadratic inside an unexpanded or unrearranged form. Always expand brackets and move everything to one side first.

Example: Solve $x(x+2)=18$ . Expand: $x^2+2x=18$ . Rearrange: $x^2+2x-18=0$ ... this doesn't factorise neatly, so switch to the formula if asked, or re-check the question for nicer numbers.

05 — Explanation

How do I form and solve a quadratic from a geometry context?

Translate the diagram into algebra, simplify to $ax^2+bx+c=0$ , solve, then reject any root that doesn't make physical sense (e.g. a negative length).

Worked Example

A trapezium has parallel sides $x-4$ and $x+5$ , height $2x$ , and area $351\text{ cm}^2$ . Find $x$ .

Area formula: $\dfrac{(a+b)}{2}\times h = 351$ .
Substitute: $\dfrac{(x-4)+(x+5)}{2}\times2x = 351 \Rightarrow (2x+1)\times x = 351$ .
Expand and rearrange: $2x^2+x-351=0$ .
Solve (factorising or formula): $x=13$ or $x=-13.5$ .
Reject $x=-13.5$ since a length can't be negative.

Answer: $x=13$ .

Exam Tip

Always state explicitly that you're rejecting the invalid root and why — this is often worth its own mark in context-based questions.

06 — Explanation

How do method and accuracy marks get split on this topic?

Examiners separate credit into method marks (for correctly setting up the standard form, substituting into the formula, or applying the Null Factor Law) and accuracy marks (for the final correct value). This means a student who rearranges and substitutes correctly but makes one slip calculating the discriminant can still pick up most of the available marks — but a student who skips straight to an answer with no working, even if correct, risks losing marks for insufficient method. Always show the rearrangement to $ax^2+bx+c=0$ explicitly, and write out the full substitution into the quadratic formula before simplifying, rather than jumping straight to a calculator answer.