Quadratic in disguise

In 8 of 20 recent exams
1% of the marks
Paper 1 or 2
5 to 7 marks

How it is set in the exam

Mostly Paper 1 (no calculator), 5–7 marks, often late in Section A or in Section B. The commonest shape is two parts: "show that" the equation becomes a given quadratic, then "hence solve". It also appears on Paper 2 (calculator), where the GDC does the final trig step (a markscheme also accepted solving the original equation from a graph).

Spot it in the exam

Exam phrases to spot (1)

“solve…”

In disguise (2)

the square isn't written as a square ( $4^x$ , $9^x$ , $2^{2x+1}$ ), one term needs an identity first ( $\cos^2 x$ or $\cos 2x$ next to $\sin x$ ), or a negative power has to be cleared first ( $e^{-x}$ , $\frac{1}{x}$ ).

an inequality in $\ln x$ , often a quadratic inequality from an earlier part with $\ln x$ in place of $x$ (a composite such as $f(\ln x)$ ): solve it for $u = \ln x$ as a quadratic inequality, then undo the log at each end. $e^x$ is increasing, so the signs stay the same: $-1 \le \ln x \le 2 \Rightarrow e^{-1} \le x \le e^2$ .
an exact ratio of an angle that is not standard, such as $\tan 15°$ : put the angle into an identity from an earlier part (one linking $\sin 2\theta$ with $\tan\theta$ , say) at the standard double angle, and the ratio satisfies a quadratic. Solve it, then reject the root outside the ratio's range for that angle ( $0 < \tan\theta < 1$ when $0° < \theta < 45°$ ).

Looks like it, but isn't (3)

two exponentials and no constant term, such as $e^{2x} = 6e^x$ or $4^x = 3 \times 2^x$ : every exponential collects into one, so there is no quadratic. Divide by the exponential (never $0$ ) and take logs: Solve an exponential equation (unknown in the power, or in the base). Asked as a graph's $x$ -intercept, it is Exponential and log graphs: intercepts, asymptotes, domain, sign, range.
one power of $e^x$ and a constant, such as $7 - 3e^{-x} = 0$ : isolate the power and take $\ln$ , Solve an exponential equation (unknown in the power, or in the base)
two trig terms with a common factor and no constant, such as $\sin^2 x = \sin x$ or $\sin 2x\cos x = \sin x$ : take the factor out and never divide by it; or squares of $\sin$ and $\cos$ with a constant but no first power, such as $3\sin^2\theta = 1 + \cos^2\theta$ : one identity, then $\pm$ . Both are Solve a trig equation on an interval

The key signal is a term and its square in the same equation:

$e^{2x}$ and $e^x$
It's a quadratic in
$u = e^x$
Because
$e^{2x} = (e^x)^2$
$4^x$ and $2^x$
It's a quadratic in
$u = 2^x$
Because
$4^x = (2^2)^x = (2^x)^2$
$9^x$ and $3^x$
It's a quadratic in
$u = 3^x$
Because
$9^x = (3^x)^2$
$2^{x+1}$ or $2^{2x+1}$
It's a quadratic in
$u = 2^x$
Because
$2^{x+1} = 2 \cdot 2^x$ , $\;2^{2x+1} = 2(2^x)^2$
$e^x$ and $e^{-x}$
It's a quadratic in
$u = e^x$
Because
multiply through by $e^x$ : $e^{-x} \cdot e^x = 1$
$x$ and $\frac{1}{x}$
It's a quadratic in
$u = x$
Because
multiply through by $x$ ( $x \ne 0$ )
$\sin^2 x$ and $\sin x$
It's a quadratic in
$u = \sin x$
Because
$\cos^2 x$ and $\sin x$
It's a quadratic in
$u = \sin x$
Because
after $\cos^2 x = 1 - \sin^2 x$
$\cos 2x$ and $\sin x$
It's a quadratic in
$u = \sin x$
Because
after $\cos 2x = 1 - 2\sin^2 x$
$\cos 2x$ and $\cos x$
It's a quadratic in
$u = \cos x$
Because
after $\cos 2x = 2\cos^2 x - 1$
$x$ and $\sqrt{x}$
It's a quadratic in
$u = \sqrt{x}$
Because
$x = (\sqrt{x})^2$
$x^4$ and $x^2$
It's a quadratic in
$u = x^2$
Because

You see	It's a quadratic in	Because
$e^{2x}$ and $e^x$	$u = e^x$	$e^{2x} = (e^x)^2$
$4^x$ and $2^x$	$u = 2^x$	$4^x = (2^2)^x = (2^x)^2$
$9^x$ and $3^x$	$u = 3^x$	$9^x = (3^x)^2$
$2^{x+1}$ or $2^{2x+1}$	$u = 2^x$	$2^{x+1} = 2 \cdot 2^x$ , $\;2^{2x+1} = 2(2^x)^2$
$e^x$ and $e^{-x}$	$u = e^x$	multiply through by $e^x$ : $e^{-x} \cdot e^x = 1$
$x$ and $\frac{1}{x}$	$u = x$	multiply through by $x$ ( $x \ne 0$ )
$\sin^2 x$ and $\sin x$	$u = \sin x$
$\cos^2 x$ and $\sin x$	$u = \sin x$	after $\cos^2 x = 1 - \sin^2 x$
$\cos 2x$ and $\sin x$	$u = \sin x$	after $\cos 2x = 1 - 2\sin^2 x$
$\cos 2x$ and $\cos x$	$u = \cos x$	after $\cos 2x = 2\cos^2 x - 1$
$x$ and $\sqrt{x}$	$u = \sqrt{x}$	$x = (\sqrt{x})^2$
$x^4$ and $x^2$	$u = x^2$

Worked example

Exponential

Paper 1 (no calculator)

Question

Solve $e^{2x} - 5e^x + 6 = 0$ , giving your answers exactly. [5]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

Try one like it

Steps

Spot the term and its square. If needed, rewrite first (index laws, an identity, or multiply through).
Substitute $u$ and write a quadratic in $u$ equal to $0$ .
Solve for $u$ .
Substitute back and solve for $x$ .
Reject impossible values, and say why:
$e^x > 0$ , $a^x > 0$ , $-1 \leq \sin x \leq 1$ , $-1 \le \cos x \le 1$ , $\sqrt{x} \geq 0$ , $x^2 \ge 0$ . The reason earns the ; a rejected value still sitting in the final answer also costs the final .
For trig, find every solution in the interval, in the units the interval uses (and none outside it).

On , show every stage (the substitution, the quadratic in $u$ solved, the back-substitution): an answer with no working can score 0, even when it is right.

Common mistakes

Stopping at $u$ .
The question asks for $x$ .
Not rejecting
$e^x = -1$ or $\sin x = 2$ , or rejecting without saying why (loses the ).
Leaving a rejected value in the answer
an extra angle outside the interval, or a second real value that breaks a condition, often loses the final ; an impossible value such as $\ln(-2)$ left beside the right answer loses the for rejecting it. The final line lists only the values that work.

More mistakes (4)

Answers with no working
on can score 0, even when they are right. Write the quadratic in $u$ and how you solved it.
Missing trig solutions
$\sin x = \frac{1}{2}$ has two solutions in $[0, 2\pi)$ . Sketch the graph or use the unit circle.
Degree answers when the interval is in radians
(or the reverse). $0 \le x < 2\pi$ means radians: $\frac{\pi}{6}$ , not $30°$ .
Writing $4^x = 2 \cdot 2^x$ .
It is $(2^x)^2$ .

Easy to confuse with

Next typeFind the equation from given features

Now try one like it

Keep going

Easy to confuse with

More in Quadratics