AS & A Level · Practice questions · Equations quadratic in a function of x

Substitute, solve, undo.

Six original Cambridge-style questions on equations that are quadratic in a function of \(x\), ending with a geometry problem that hides a quartic.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Find \(u\), solve for \(u\), then find \(x\).

Spot two terms where one power is double the other.
Substitute \(u\) for the smaller power (or the function) and solve the quadratic in \(u\).
Reject values the function can never take: negative \(x^{2}\), negative \(\sqrt{x}\), \(\sin x\) outside \([-1, 1]\).
Undo the substitution for each remaining value, including both signs for \(x^{2} = k\).
Check in the original equation when a square root is involved.

Recall

[3 marks]

Solve \(x^{4} - 10x^{2} + 9 = 0\).

Let \(u = x^{2}\): \(u^{2} - 10u + 9 = 0 \Rightarrow (u - 1)(u - 9) = 0\). ✓
\(x^{2} = 1\) or \(x^{2} = 9\). ✓
\(x = \pm 1\) or \(x = \pm 3\). ✓

Calculation

[3 marks]

Solve \(x - 6\sqrt{x} + 8 = 0\).

Let \(u = \sqrt{x}\): \(u^{2} - 6u + 8 = 0 \Rightarrow (u - 2)(u - 4) = 0\). ✓
\(\sqrt{x} = 2\) or \(\sqrt{x} = 4\), both positive, so both valid. ✓
\(x = 4\) or \(x = 16\). ✓

Calculation

[3 marks]

Find the real values of \(x\) for which \(4x^{4} + 7x^{2} - 2 = 0\).

Let \(u = x^{2}\): \(4u^{2} + 7u - 2 = 0 \Rightarrow (4u - 1)(u + 2) = 0\). ✓
\(x^{2} = -2\) has no real solutions, so reject it. ✓
\(x^{2} = \dfrac{1}{4} \Rightarrow x = \pm\dfrac{1}{2}\). ✓

Calculation

[3 marks]

Solve \(\dfrac{6}{x^{2}} - \dfrac{1}{x} - 1 = 0\).

Let \(u = \dfrac{1}{x}\): \(6u^{2} - u - 1 = 0 \Rightarrow (3u + 1)(2u - 1) = 0\). ✓
\(u = \dfrac{1}{2} \Rightarrow x = 2\). ✓
\(u = -\dfrac{1}{3} \Rightarrow x = -3\). ✓

Analysis

[4 marks]

Find the coordinates of the points where the curves \(y = x^{2} + 2\) and \(y = \dfrac{3}{x^{2}}\) intersect.

\(x^{2} + 2 = \dfrac{3}{x^{2}} \Rightarrow x^{4} + 2x^{2} - 3 = 0\). ✓
Let \(u = x^{2}\): \((u + 3)(u - 1) = 0\); \(u = -3\) is impossible. ✓
\(x^{2} = 1 \Rightarrow x = \pm 1\), and \(y = 1 + 2 = 3\). ✓
The curves meet at \((1, 3)\) and \((-1, 3)\). ✓

Analysis

[5 marks]

A right-angled triangle has area 2 cm². One of the shorter sides is \(x\) cm and the hypotenuse is \(\sqrt{17}\) cm.

(a) Show that the other shorter side is \(\dfrac{4}{x}\) cm and that \(x^{4} - 17x^{2} + 16 = 0\).

(b) Find the lengths of the two shorter sides.

(a) Area \(= \dfrac{1}{2} \times x \times y = 2 \Rightarrow y = \dfrac{4}{x}\). ✓
Pythagoras: \(x^{2} + \dfrac{16}{x^{2}} = 17\); multiply by \(x^{2}\): \(x^{4} - 17x^{2} + 16 = 0\). ✓
(b) Let \(u = x^{2}\): \((u - 1)(u - 16) = 0\), so \(x^{2} = 1\) or \(16\). ✓
\(x\) is a length, so \(x > 0\): \(x = 1\) or \(x = 4\). ✓
Either way the shorter sides are 1 cm and 4 cm (check: \(1 + 16 = 17\), area \(\frac{1}{2} \times 1 \times 4 = 2\)). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.