Home / A-Level Maths 9709 / Quadratics / Quadratic equations and inequalities
Cambridge International AS & A Level · Topic 1.1
A-Level Maths 9709 / Topic 1.1 / AS & A Level

Find the roots, then sketch.

Solve quadratic equations by factorising, completing the square or the formula, then use the roots and a quick sketch to solve quadratic inequalities, including the sets of values that discriminant questions produce.

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The key idea

To solve a quadratic inequality, first solve the equation to find the critical values, then sketch the parabola. For \(a > 0\): \((x - \alpha)(x - \beta) < 0\) gives \(\alpha < x < \beta\) (between the roots); \((x - \alpha)(x - \beta) > 0\) gives \(x < \alpha\) or \(x > \beta\) (outside them).

x y −3 5 y < 0 here: −3 < x < 5 y > 0 y > 0 y = (x + 3)(x − 5)
Fig. 1: y = x² - 2x - 15 = (x + 3)(x - 5) is below the x-axis between the roots, so x² - 2x - 15 < 0 for -3 < x < 5
Section 01

Three ways to solve.

In the lab, pick a quadratic and solve it by factorising, then by the formula, and check that both routes land on the same roots. Always rearrange to \(ax^{2} + bx + c = 0\) first. Factorise when it is quick; otherwise use the formula \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) or complete the square.

Worked example Factorising
Solve \(6x^{2} - x - 2 = 0\).
  1. Look for two numbers with product \(6 \times (-2) = -12\) and sum \(-1\): \(-4\) and \(3\).
  2. \(6x^{2} - 4x + 3x - 2 = 2x(3x - 2) + (3x - 2) = (3x - 2)(2x + 1)\).
  3. Set each factor to zero: \(3x - 2 = 0\) or \(2x + 1 = 0\).
Answer: \(x = \dfrac{2}{3}\) or \(x = -\dfrac{1}{2}\)
Worked example The formula
Solve \(3x^{2} - 7x + 1 = 0\), giving your answers exactly and to 3 significant figures.
  1. \(a = 3,\ b = -7,\ c = 1\), so \(b^{2} - 4ac = 49 - 12 = 37\).
  2. \(x = \dfrac{7 \pm \sqrt{37}}{6}\).
  3. \(\sqrt{37} = 6.0827...\), so \(x = \dfrac{13.0827...}{6}\) or \(\dfrac{0.9172...}{6}\).
Answer: \(x = \dfrac{7 \pm \sqrt{37}}{6}\), i.e. \(x = 2.18\) or \(x = 0.153\)
Worked example Rearrange first
Solve \(x(2x - 5) = 3\).
  1. A product equal to 3 tells you nothing, so expand and rearrange: \(2x^{2} - 5x - 3 = 0\).
  2. Factorise: \((2x + 1)(x - 3) = 0\).
Answer: \(x = -\dfrac{1}{2}\) or \(x = 3\)
Examiner trap

Never divide both sides by \(x\): \(x^{2} = 4x\) becomes \(x = 4\) and the root \(x = 0\) is lost. Rearrange to \(x^{2} - 4x = 0\), so \(x(x - 4) = 0\) and \(x = 0\) or \(x = 4\).

Section 02

Inequalities: sketch first.

A quadratic inequality is a question about the graph: where is the parabola above, or below, the \(x\)-axis? Find the critical values from the equation, sketch the curve through them, and read the answer off the sketch (Fig. 1).

Worked example Between the roots
Solve \(x^{2} - 2x - 15 < 0\).
  1. Critical values: \(x^{2} - 2x - 15 = (x + 3)(x - 5) = 0\), so \(x = -3\) or \(x = 5\).
  2. The parabola opens upwards and is below the \(x\)-axis between its roots.
  3. For \(\ge 0\) instead, the answer would be the outside: \(x \le -3\) or \(x \ge 5\).
Answer: \(-3 < x < 5\)
Worked example Outside the roots
Solve \(2x^{2} + 5x > 3\).
  1. Rearrange so one side is zero: \(2x^{2} + 5x - 3 > 0\).
  2. Factorise: \((2x - 1)(x + 3) > 0\), critical values \(x = \dfrac{1}{2}\) and \(x = -3\).
  3. The curve is above the axis outside the roots.
Answer: \(x < -3\) or \(x > \dfrac{1}{2}\)
Worked example Negative coefficient of x²
Solve \(12 + x - x^{2} \ge 0\).
  1. Multiply by \(-1\) and reverse the sign: \(x^{2} - x - 12 \le 0\).
  2. \((x - 4)(x + 3) \le 0\), critical values \(4\) and \(-3\).
  3. Below or on the axis means between the roots, ends included.
Answer: \(-3 \le x \le 4\)
Examiner trap

\(x^{2} > 9\) does NOT mean \(x > \pm 3\). The curve \(y = x^{2} - 9\) is above the axis outside \(-3\) and \(3\), so \(x < -3\) or \(x > 3\).

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 \(x^{2} - 7x + 12 = 0\).

Quick check+10 XP

Solve \(x^{2} < 16\).

Quick check+10 XP

Solve \((x - 1)(x - 6) > 0\).

Quick check+10 XP

Solve \(2x^{2} - 3x - 2 = 0\).

Section 03

Surds, constants and sets of values.

When the quadratic does not factorise, find the critical values with the formula and keep them exact. The same method finishes every discriminant question that asks for a set of values of \(k\).

Worked example Irrational critical values
Solve \(x^{2} - 4x - 1 < 0\), leaving your answer in surd form.
  1. \(x^{2} - 4x - 1 = (x - 2)^{2} - 5 = 0\) gives \(x = 2 \pm \sqrt{5}\).
  2. The curve is below the axis between the roots.
Answer: \(2 - \sqrt{5} < x < 2 + \sqrt{5}\)
Worked example A set of values of k
Find the set of values of \(k\) for which \(x^{2} + kx + k + 3 = 0\) has two distinct real roots.
  1. Two distinct roots: \(k^{2} - 4(k + 3) > 0\), so \(k^{2} - 4k - 12 > 0\).
  2. \((k - 6)(k + 2) > 0\), critical values \(6\) and \(-2\).
  3. A positive quadratic in \(k\) is greater than zero outside its roots.
Answer: \(k < -2\) or \(k > 6\)
Your turn
Solve \(x(x + 3) \le 10\).
Examiner trap

An 'outside' answer is two separate pieces joined by 'or'. Writing \(6 < k < -2\) is meaningless (no number is both bigger than 6 and smaller than \(-2\)) and loses the mark.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Solve the inequality \(3x^{2} \le 2 - 5x\).

Exam style+20 XP

Solve \(4 + 3x - x^{2} < 0\).

Exam style+20 XP

Solve \(2x^{2} - 6x + 1 = 0\), giving your answers in exact form.

Exam style+20 XP

Find the set of values of \(k\) for which \(x^{2} + (k + 1)x + 4 = 0\) has two distinct real roots.

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Equations solved, inequalities sketched.

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Go deeper · practice
Six original Cambridge-style questions
Factorising, the formula to 3 s.f., inequalities between and outside the roots, an inequality that simplifies to surds, a set of values of \(k\), and the size of a rectangle.