Home / IGCSE Add Maths 0606 / Equations, inequalities and graphs / Equations in quadratic form
Cambridge IGCSE · Topic 4.3
IGCSE Add Maths 0606 / Topic 4 / IGCSE

A quadratic in disguise.

Equations such as \(x^{4} - 13x^{2} + 36 = 0\), \(x - 2\sqrt{x} - 8 = 0\) and \(e^{x} + 6e^{-x} = 5\) are quadratics wearing a mask. Spot the expression that appears squared, call it \(u\), solve the quadratic in \(u\), then undo the substitution and throw out any value that is impossible.

Topic 4All equations, inequalities and graphs lessons →
The key idea

An equation is in quadratic form when it can be written \(au^{2} + bu + c = 0\) with \(u\) an expression in \(x\): look for one term that is the square of another, such as \(x^{4} = (x^{2})^{2}\), \(x = (\sqrt{x})^{2}\), \(e^{2x} = (e^{x})^{2}\) or \(4^{x} = (2^{x})^{2}\). Clear any \(e^{-x}\) by multiplying every term by \(e^{x}\). Solve for \(u\), then solve \(u = \) each value for \(x\), rejecting the impossible ones: \(e^{x} > 0\), \(2^{x} > 0\), \(\sqrt{x} \ge 0\) and \(x^{2} \ge 0\). A value of \(\ln x\) may be negative; only the argument of \(\ln\) must be positive.

x y O 1 2 3 1 (0, 2) ln 2 ln 3 y = e2x − 5ex + 6 let u = ex, so e2x = u² u² − 5u + 6 = 0 (u − 2)(u − 3) = 0 ex = 2 or ex = 3 x = ln 2 or x = ln 3 x = 0: y = 1 − 5 + 6 = 2
Fig. 1: the curve y = e^(2x) - 5e^x + 6. With u = e^x the equation y = 0 becomes u² - 5u + 6 = 0, so e^x = 2 or 3, and the curve crosses the x-axis at x = ln 2 and x = ln 3. It meets the y-axis at (0, 2)
Section 01

Spot the hidden square.

The trick in every question of this lesson is the same: find an expression \(u\) so that the equation contains only \(u^{2}\), \(u\) and a constant. The syllabus expects you to choose the substitution yourself, so learn to recognise the pairs below. Write \(u\) down clearly, solve the quadratic in \(u\) by factorising or the formula, and then go back to \(x\): the answers to the quadratic are values of \(u\), not of \(x\).

Terms in the equationSubstitutionThe square
\(x^{4}\) and \(x^{2}\)\(u = x^{2}\)\(x^{4} = u^{2}\)
\(x\) and \(\sqrt{x}\)\(u = \sqrt{x}\)\(x = u^{2}\)
\(x^{\frac{2}{3}}\) and \(x^{\frac{1}{3}}\)\(u = x^{\frac{1}{3}}\)\(x^{\frac{2}{3}} = u^{2}\)
\(e^{2x}\) and \(e^{x}\)\(u = e^{x}\)\(e^{2x} = u^{2}\)
\(2^{2x}\) (or \(4^{x}\)) and \(2^{x}\)\(u = 2^{x}\)\(2^{2x} = 4^{x} = u^{2}\)
\((\ln x)^{2}\) and \(\ln x\)\(u = \ln x\)\((\ln x)^{2} = u^{2}\)
Worked example A quartic that is really a quadratic
Solve \(x^{4} - 13x^{2} + 36 = 0\).
  1. \(x^{4} = (x^{2})^{2}\), so let \(u = x^{2}\): \(u^{2} - 13u + 36 = 0\).
  2. Factorise: \((u - 4)(u - 9) = 0\), so \(u = 4\) or \(u = 9\).
  3. Undo: \(x^{2} = 4\) gives \(x = \pm 2\); \(x^{2} = 9\) gives \(x = \pm 3\).
  4. A quartic can have up to four roots, and this one has all four.
Answer: \(x = -3,\ -2,\ 2\) or \(3\)
Worked example Fractional powers
Solve \(x^{\frac{2}{3}} + x^{\frac{1}{3}} - 12 = 0\).
  1. \(x^{\frac{2}{3}} = \left(x^{\frac{1}{3}}\right)^{2}\), so let \(u = x^{\frac{1}{3}}\): \(u^{2} + u - 12 = 0\).
  2. Factorise: \((u + 4)(u - 3) = 0\), so \(u = 3\) or \(u = -4\).
  3. Undo by cubing: \(x^{\frac{1}{3}} = 3\) gives \(x = 3^{3} = 27\); \(x^{\frac{1}{3}} = -4\) gives \(x = (-4)^{3} = -64\).
  4. A cube root can be negative, so keep both. Check \(x = -64\): \(16 + (-4) - 12 = 0\).
Answer: \(x = 27\) or \(x = -64\)
Your turn
Solve \(x^{4} - 5x^{2} + 4 = 0\).
Examiner trap

Do not stop at the values of \(u\). In \(x^{4} - 13x^{2} + 36 = 0\) the answer is not '\(4\) and \(9\)': those are values of \(x^{2}\). And \(x^{2} = 4\) has TWO solutions, \(x = 2\) and \(x = -2\); forgetting the negative root loses half the answers.

Section 02

Roots and impossible values.

Some values of \(u\) cannot happen. \(\sqrt{x}\) means the positive square root, so \(\sqrt{x} \ge 0\). A square \(x^{2}\) is never negative. A power such as \(2^{x}\) is always positive. When the quadratic gives a value of \(u\) that breaks one of these rules, reject it and say why: that is a mark in the exam.

Worked example A square root that must be positive
Solve \(x - 2\sqrt{x} - 8 = 0\).
  1. \(x = (\sqrt{x})^{2}\), so let \(u = \sqrt{x}\): \(u^{2} - 2u - 8 = 0\).
  2. Factorise: \((u - 4)(u + 2) = 0\), so \(u = 4\) or \(u = -2\).
  3. \(\sqrt{x} = -2\) is impossible, because \(\sqrt{x} \ge 0\). Reject it.
  4. \(\sqrt{x} = 4\) gives \(x = 16\). Check: \(16 - 2(4) - 8 = 0\).
Answer: \(x = 16\) only
Worked example Powers of 2
Solve \(4^{x} - 3(2^{x}) - 4 = 0\).
  1. \(4^{x} = (2^{2})^{x} = (2^{x})^{2}\), so let \(u = 2^{x}\): \(u^{2} - 3u - 4 = 0\).
  2. Factorise: \((u - 4)(u + 1) = 0\), so \(u = 4\) or \(u = -1\).
  3. \(2^{x} = -1\) is impossible, since \(2^{x} > 0\) for every \(x\). Reject it.
  4. \(2^{x} = 4 = 2^{2}\), so \(x = 2\).
Answer: \(x = 2\)
Your turn
Solve \(x^{4} + 2x^{2} - 8 = 0\), giving exact answers.
Examiner trap

Squaring \(\sqrt{x} = -2\) gives \(x = 4\), but \(x = 4\) does NOT satisfy the equation: \(4 - 2(2) - 8 = -8\). Reject the impossible value of \(u\) before you undo the substitution, and if in doubt, check each answer in the ORIGINAL equation.

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

Which substitution turns \(x^{\frac{2}{3}} - 5x^{\frac{1}{3}} + 6 = 0\) into a quadratic?

Quick check+10 XP

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

Quick check+10 XP

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

Quick check+10 XP

Solve \(x - 3\sqrt{x} - 4 = 0\).

Section 03

Exponentials and logarithms.

With \(e^{x}\), the inverse is the natural logarithm: \(e^{x} = k\) gives \(x = \ln k\), which exists only when \(k > 0\). If the equation contains \(e^{-x}\), remember \(e^{-x} = \dfrac{1}{e^{x}}\) and multiply EVERY term by \(e^{x}\) to clear it. With \(u = \ln x\) it works the other way: \(\ln x = k\) gives \(x = e^{k}\), and here \(k\) may be negative. Paper 1 has no calculator, so leave answers exact, as \(\ln 3\) or \(e^{2}\).

Worked example Clear the negative power
Solve \(e^{x} + 6e^{-x} = 5\), giving exact answers.
  1. Multiply every term by \(e^{x}\): \(e^{2x} + 6 = 5e^{x}\), since \(e^{-x} \times e^{x} = 1\).
  2. Rearrange and let \(u = e^{x}\): \(u^{2} - 5u + 6 = 0\).
  3. Factorise: \((u - 2)(u - 3) = 0\), so \(e^{x} = 2\) or \(e^{x} = 3\). Both are positive, so both are allowed.
  4. Take \(\ln\): \(x = \ln 2\) or \(x = \ln 3\). These are where the curve in Fig. 1 crosses the \(x\)-axis.
Answer: \(x = \ln 2\) or \(x = \ln 3\)
Worked example A squared logarithm
Solve \(2(\ln 5x)^{2} + \ln 5x - 6 = 0\), giving exact answers.
  1. Let \(u = \ln 5x\): \(2u^{2} + u - 6 = 0\).
  2. Factorise: \((2u - 3)(u + 2) = 0\), so \(u = \dfrac{3}{2}\) or \(u = -2\).
  3. A logarithm can be negative, so keep both. \(\ln 5x = \dfrac{3}{2}\) gives \(5x = e^{\frac{3}{2}}\), so \(x = \dfrac{e^{\frac{3}{2}}}{5}\).
  4. \(\ln 5x = -2\) gives \(5x = e^{-2}\), so \(x = \dfrac{1}{5e^{2}}\). Both values of \(x\) are positive, so \(\ln 5x\) exists.
Answer: \(x = \dfrac{e^{\frac{3}{2}}}{5}\) or \(x = \dfrac{1}{5e^{2}}\)
Your turn
Solve \(e^{2x} + e^{x} - 6 = 0\), giving your answer in exact form.
Examiner trap

When you multiply by \(e^{x}\), the constant term gets multiplied too: \(e^{x} - 2e^{-x} = 1\) becomes \(e^{2x} - 2 = e^{x}\), not \(e^{2x} - 2 = 1\). Reject \(e^{x} < 0\) (there is no \(\ln\) of a negative number), but do NOT reject a negative value of \(\ln x\): \(\ln x = -2\) just means \(x = e^{-2}\).

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

Solve \(2^{2x + 1} - 9(2^{x}) + 4 = 0\).

Exam style+20 XP

Solve \(e^{x} - 2e^{-x} = 1\).

Exam style+20 XP

Solve \(2(\ln x)^{2} - 5\ln x + 2 = 0\).

Exam style+20 XP

Solve \(6x + 7\sqrt{x} - 3 = 0\).

Skill unlocked

Mask off: the quadratic was there all along.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
A quartic in \(x^{2}\), fractional powers, \(3e^{x} = 12 - 5e^{-x}\) with surd answers, a squared logarithm, powers of 2 with a rejected root and a curve \(y = e^{2x} - 5e^{x} + 6\).