IGCSE · Practice questions · Equations in quadratic form

Substitute, solve, undo.

Six original Cambridge-style questions on equations in quadratic form: choosing the substitution, solving the quadratic, undoing it, rejecting impossible values and giving exact answers with \(\ln\) and \(e\).

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

The method

Spot \(u\), solve, undo, reject.

Spot the square: find \(u\) so that the equation has only \(u^{2}\), \(u\) and a constant; rewrite \(4^{x}\), \(2^{x + 2}\) or \(e^{-x}\) first.
Clear negative powers: multiply every term, constant included, by \(e^{x}\).
Solve the quadratic in \(u\) by factorising or the formula.
Reject impossible values: \(e^{x} > 0\), \(2^{x} > 0\), \(\sqrt{x} \ge 0\), \(x^{2} \ge 0\). A negative \(\ln x\) is fine.
Undo the substitution for each value left and give exact answers (\(\ln k\), \(e^{k}\), surds).

Recall

[3 marks]

Solve the equation \(x^{4} - 17x^{2} + 16 = 0\).

Let \(u = x^{2}\): \(u^{2} - 17u + 16 = 0\). ✓
\((u - 1)(u - 16) = 0\), so \(x^{2} = 1\) or \(x^{2} = 16\). ✓
\(x = -4,\ -1,\ 1\) or \(4\). ✓

Calculation

[4 marks]

Using a suitable substitution, solve the equation \(x^{\frac{2}{3}} - 3x^{\frac{1}{3}} - 4 = 0\).

Let \(u = x^{\frac{1}{3}}\), so \(x^{\frac{2}{3}} = u^{2}\): \(u^{2} - 3u - 4 = 0\). ✓
\((u - 4)(u + 1) = 0\), so \(x^{\frac{1}{3}} = 4\) or \(x^{\frac{1}{3}} = -1\). ✓
Cube: \(x = 64\) or \(x = -1\). A cube root may be negative, so both are valid. ✓
Check \(x = -1\): \(1 - 3(-1) - 4 = 0\). ✓

Calculation

[5 marks]

Solve the equation \(3e^{x} = 12 - 5e^{-x}\), giving your answers in the form \(\ln\left(\dfrac{a \pm \sqrt{b}}{c}\right)\), where \(a\), \(b\) and \(c\) are integers.

Multiply every term by \(e^{x}\): \(3e^{2x} = 12e^{x} - 5\). ✓
Let \(u = e^{x}\): \(3u^{2} - 12u + 5 = 0\). ✓
\(u = \dfrac{12 \pm \sqrt{144 - 60}}{6} = \dfrac{12 \pm \sqrt{84}}{6} = \dfrac{12 \pm 2\sqrt{21}}{6} = \dfrac{6 \pm \sqrt{21}}{3}\). ✓
\(\sqrt{21}\) is between 4 and 5, so \(6 - \sqrt{21} > 0\): both values of \(e^{x}\) are positive and both are allowed. ✓
\(x = \ln\left(\dfrac{6 + \sqrt{21}}{3}\right)\) or \(x = \ln\left(\dfrac{6 - \sqrt{21}}{3}\right)\). ✓

Analysis

[4 marks]

Solve the equation \(3(\ln 2x)^{2} - 5\ln 2x - 2 = 0\), giving your answers in exact form.

Let \(u = \ln 2x\): \(3u^{2} - 5u - 2 = 0\). ✓
\((3u + 1)(u - 2) = 0\), so \(\ln 2x = 2\) or \(\ln 2x = -\dfrac{1}{3}\). A negative logarithm is allowed. ✓
\(2x = e^{2}\), so \(x = \dfrac{e^{2}}{2}\). ✓
\(2x = e^{-\frac{1}{3}}\), so \(x = \dfrac{1}{2}e^{-\frac{1}{3}}\). ✓

Analysis

[4 marks]

Solve the equation \(2^{2x} - 2^{x + 2} - 32 = 0\), explaining why one value of your substitution is rejected.

\(2^{x + 2} = 2^{2} \times 2^{x} = 4(2^{x})\). Let \(u = 2^{x}\): \(u^{2} - 4u - 32 = 0\). ✓
\((u - 8)(u + 4) = 0\), so \(2^{x} = 8\) or \(2^{x} = -4\). ✓
\(2^{x} > 0\) for every real \(x\), so \(2^{x} = -4\) has no solution: reject it. ✓
\(2^{x} = 8 = 2^{3}\), so \(x = 3\). ✓

Proof

[6 marks]

The curve \(y = e^{2x} - 5e^{x} + 6\) is shown in Fig. 1 of the lesson.

(a) Find the coordinates of the point where the curve meets the \(y\)-axis.

(b) Show that the curve meets the \(x\)-axis at \(x = \ln 2\) and \(x = \ln 3\).

(a) At \(x = 0\): \(y = e^{0} - 5e^{0} + 6 = 1 - 5 + 6 = 2\), so the point is \((0, 2)\). ✓
(b) Set \(y = 0\) and let \(u = e^{x}\): \(u^{2} - 5u + 6 = (u - 2)(u - 3) = 0\). ✓
So \(e^{x} = 2\) or \(e^{x} = 3\), both positive, giving \(x = \ln 2\) or \(x = \ln 3\). ✓
(c) \(u^{2} - 5u - 6 = (u - 6)(u + 1) = 0\), so \(e^{x} = 6\) or \(e^{x} = -1\). ✓
\(e^{x} > 0\) for every \(x\), so \(e^{x} = -1\) is impossible. ✓
The only solution is \(x = \ln 6\). ✓

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.