Solve quadratic equations by factorising, by the quadratic formula and by completing the square, choose the quickest method, give exact answers as simplified surds for the non-calculator paper, and rearrange fractions and context problems into \(ax^{2} + bx + c = 0\) first.
Rearrange to \(ax^{2} + bx + c = 0\) first. Factorise if you can; otherwise use \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) or complete the square. On Paper 1 (no calculator) leave the roots exact, with every surd simplified and every term divided through. The real roots are where \(y = ax^{2} + bx + c\) crosses the \(x\)-axis.
Fig. 1: y = x² - 2x - 2 does not factorise, but completing the square gives (x - 1)² - 3, so the curve crosses the x-axis at the exact roots 1 ± √3, an equal distance either side of the line of symmetry x = 1
Section 01
Choosing a method.
In the lab, set \(a = 1\), \(b = -2\), \(c = -2\) and read where the curve crosses the \(x\)-axis: about \(-0.73\) and \(2.73\), the decimals of \(1 \pm \sqrt{3}\) in Fig. 1. Then try \(a = 6\), \(b = -1\), \(c = -15\): the roots \(-1.5\) and \(1.67\) are rational, a sign that the quadratic factorises.
Factorise when the numbers are friendly. If \(AB = 0\) then \(A = 0\) or \(B = 0\), so one side must be zero first.
Complete the square when \(a = 1\) and \(b\) is even, or when the question asks for the form \(p \pm \sqrt{q}\): \(x^{2} + bx + c = \left(x + \tfrac{b}{2}\right)^{2} - \tfrac{b^{2}}{4} + c\).
The formula \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) (given in the List of formulas) always works. If \(b^{2} - 4ac < 0\) there are no real roots.
On Paper 2 a correct answer from your calculator earns the marks without working, but a wrong one earns nothing, so show the method anyway. Give decimals to 3 significant figures unless told otherwise.
Worked example Factorise with a not equal to 1
Solve \(6x^{2} - x - 15 = 0\).
\(ac = -90\). Two numbers with product \(-90\) and sum \(-1\): \(-10\) and \(9\).
\(ac = -24\); numbers \(8\) and \(-3\). \(2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4) = 0\), so \(x = \dfrac{3}{2}\) or \(x = -4\).
Examiner trap
The zero is what makes factorising work. \((x - 2)(x + 1) = 4\) does NOT mean \(x - 2 = 4\) or \(x + 1 = 4\): expand, subtract 4 and factorise \(x^{2} - x - 6 = (x - 3)(x + 2) = 0\). Taking the square root in \((x - 3)^{2} = 12\) also needs \(\pm\), or you lose a root.
Section 02
Exact answers in surd form.
Paper 1 has no calculator, so the formula and completing the square leave you with surds. An answer is only exact AND finished when each surd is simplified and the fraction is cancelled through every term.
Simplify a surd by taking out the largest square factor: \(\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}\), \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\).
Then divide every term of the numerator by the common factor: \(\dfrac{4 \pm 2\sqrt{10}}{6} = \dfrac{2 \pm \sqrt{10}}{3}\).
\(\sqrt{a + b}\) is NOT \(\sqrt{a} + \sqrt{b}\); only products split: \(\sqrt{ab} = \sqrt{a}\sqrt{b}\).
Worked example Formula, then simplify
Solve \(3x^{2} - 4x - 2 = 0\), giving your answers in exact form.
By completing the square, solve \(x^{2} + 10x + 7 = 0\), giving your answers in the form \(p \pm q\sqrt{2}\).
\((x + 5)^{2} - 25 + 7 = 0\), so \((x + 5)^{2} = 18\) and \(x + 5 = \pm\sqrt{18} = \pm 3\sqrt{2}\). So \(x = -5 \pm 3\sqrt{2}\).
Examiner trap
Cancel the whole numerator, not one term. \(\dfrac{4 \pm 2\sqrt{10}}{6}\) is NOT \(4 \pm \dfrac{\sqrt{10}}{3}\) and NOT \(\dfrac{2}{3} \pm 2\sqrt{10}\). If you cannot divide every term by the same number, leave the fraction as it is.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Solve \((2x - 1)(x + 4) = 0\).
Quick check+10 XP
Simplify \(\sqrt{48}\) fully.
Quick check+10 XP
By completing the square, the roots of \(x^{2} - 8x + 5 = 0\) are
Quick check+10 XP
Solve \(x^{2} = 5x\).
Section 03
Rearrange first, then solve.
Many quadratics arrive in disguise: fractions with \(x\) in the denominator, \(x\) on both sides, or a problem in words. Clear the fractions, collect every term on one side, solve, and then check each root against the original equation and the context.
Fractions: multiply EVERY term (the right-hand side too) by the product of the denominators. A root that makes a denominator zero must be rejected.
Context: a length, a time or a number of items cannot be negative. Reject such roots and say why.
Equations such as \(e^{2x} - 5e^{x} + 6 = 0\) are quadratic in \(e^{x}\); substitutions like this come in topic 4.
Worked example Clear the fractions
Solve \(\dfrac{3}{x} + \dfrac{2}{x + 1} = 4\).
Multiply every term by \(x(x + 1)\): \(3(x + 1) + 2x = 4x(x + 1)\).
\(5x + 3 = 4x^{2} + 4x\), so \(4x^{2} - x - 3 = 0\).
\((4x + 3)(x - 1) = 0\). Neither root makes a denominator zero.
Answer: \(x = 1\) or \(x = -\dfrac{3}{4}\)
Worked example A right-angled triangle
The sides of a right-angled triangle are \(x\) cm, \((x + 7)\) cm and a hypotenuse of \((x + 8)\) cm. Find the lengths of the sides.
\((x - 5)(x + 3) = 0\), so \(x = 5\) or \(x = -3\).
A length cannot be negative, so reject \(x = -3\).
Answer: 5 cm, 12 cm and 13 cm
Your turn
Solve \((2x - 1)(x + 3) = x + 9\), giving your answers in the form \(p \pm \sqrt{q}\).
\(2x^{2} + 5x - 3 = x + 9\), so \(2x^{2} + 4x - 12 = 0\) and \(x^{2} + 2x - 6 = 0\). Then \((x + 1)^{2} = 7\), so \(x = -1 \pm \sqrt{7}\).
Examiner trap
When you clear fractions, the number on the right must be multiplied too: \(\dfrac{3}{x} + \dfrac{2}{x + 1} = 4\) becomes \(3(x + 1) + 2x = 4x(x + 1)\), not \(= 4\). And in context, finish the job: state the lengths that were asked for, not just \(x\).
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Additional Mathematics Paper 1 and Paper 2 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Without using a calculator, solve \(2x^{2} - 6x + 3 = 0\), giving your answers in exact form.
Exam style+20 XP
Solve \(\dfrac{6}{x} = x + 1\).
Exam style+20 XP
A rectangle has length \((x + 4)\) cm, width \(x\) cm and area 20 cm2. Find the exact width.
Exam style+20 XP
Solve \((x - 3)(x + 2) = 6\).
Skill unlocked
Roots found, exactly.
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