Equations that are quadratic in a function of x

AS · P1 · 9 min

Some equations are quadratics in disguise. $x^4 - 5x^2 + 4 = 0$ is a quadratic in $x^2$ ; $6x + \sqrt{x} - 2 = 0$ is a quadratic in $\sqrt{x}$ ; $\tan^2 x = 1 + \tan x$ is a quadratic in $\tan x$ . Once you spot the hidden variable, a substitution turns the problem into one you can already solve. The syllabus names this skill explicitly, and it reappears in trigonometry and (in Paper 3) with exponentials, so the habit of spotting the disguise is worth building now.

Spotting the disguise

A quadratic in $u$ has the form $au^2 + bu + c = 0$ . The equation is quadratic in some function $f(x)$ if you can write it as

a\left[f(x)\right]^2 + b\,f(x) + c = 0

The test: one of the powers is double the other, and there is a constant term (or the constant is zero).

Key result

Equation contains	Substitute	Because
$x^4$ and $x^2$	$u = x^2$	$x^4 = \left(x^2\right)^2$
$x^6$ and $x^3$	$u = x^3$	$x^6 = \left(x^3\right)^2$
$x$ and $\sqrt{x}$	$u = \sqrt{x}$	$x = \left(\sqrt{x}\right)^2$
$x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$	$u = x^{\frac{1}{3}}$	$x^{\frac{2}{3}} = \left(x^{\frac{1}{3}}\right)^2$
$\dfrac{1}{x^2}$ and $\dfrac{1}{x}$	$u = \dfrac{1}{x}$	$\dfrac{1}{x^2} = \left(\dfrac{1}{x}\right)^2$
$\tan^2 x$ and $\tan x$	$u = \tan x$	$\tan^2 x = (\tan x)^2$
$\sin^2 x$ and $\sin x$	$u = \sin x$	$\sin^2 x = (\sin x)^2$
$(x^2 + x)^2$ and $(x^2 + x)$	$u = x^2 + x$	the bracket is the hidden variable

Solving a disguised quadratic

Identify the hidden variable and write $u = \dots$ .
Rewrite the equation as a quadratic in $u$ and solve it.
Check each value of $u$ is possible. $x^2$ and $\sqrt{x}$ cannot be negative; $\sin x$ and $\cos x$ must lie between $-1$ and $1$ .
Substitute back, $u = f(x)$ , and solve for $x$ . Remember $x^2 = 4$ gives $x = \pm 2$ .
List every solution for $x$ .

Step 3 is where marks are commonly lost: an impossible value of $u$ must be rejected with a reason, for example " $x^2 = -1$ has no real solutions".

Quadratic in x squared

Solve $x^4 - 5x^2 + 4 = 0$ .

Solution

Let $u = x^2$ , so $x^4 = u^2$ :

u^2 - 5u + 4 = 0 \quad\Rightarrow\quad (u - 1)(u - 4) = 0 \quad\Rightarrow\quad u = 1 \text{ or } u = 4

Back-substitute: $x^2 = 1$ gives $x = \pm 1$ ; $x^2 = 4$ gives $x = \pm 2$ .

Solutions: $x = -2, -1, 1, 2$ .

y = x^4 - 5x^2 + 4 (-2, 0) (-1, 0) (1, 0) (2, 0)

The graph confirms four crossings of the $x$ -axis.

Rejecting an impossible value

Solve $x^4 - 3x^2 - 4 = 0$ .

Solution

Let $u = x^2$ : $u^2 - 3u - 4 = 0$ , so $(u - 4)(u + 1) = 0$ and $u = 4$ or $u = -1$ .

$x^2 = 4$ gives $x = \pm 2$ .
$x^2 = -1$ has no real solutions, since a square cannot be negative.

Solutions: $x = \pm 2$ .

Quadratic in the square root of x

Solve $6x + \sqrt{x} - 2 = 0$ .

Solution

Let $u = \sqrt{x}$ , so $x = u^2$ :

6u^2 + u - 2 = 0 \quad\Rightarrow\quad (3u + 2)(2u - 1) = 0 \quad\Rightarrow\quad u = \tfrac{1}{2} \text{ or } u = -\tfrac{2}{3}

$\sqrt{x}$ means the non-negative square root, so $\sqrt{x} = -\tfrac{2}{3}$ is impossible. From $\sqrt{x} = \tfrac{1}{2}$ , $x = \tfrac{1}{4}$ .

Check: $6 \times \tfrac{1}{4} + \tfrac{1}{2} - 2 = \tfrac{3}{2} + \tfrac{1}{2} - 2 = 0$ . Correct.

Fractional powers

Solve $2x^{\frac{2}{3}} - 7x^{\frac{1}{3}} + 3 = 0$ .

Solution

Let $u = x^{\frac{1}{3}}$ , so $x^{\frac{2}{3}} = u^2$ :

2u^2 - 7u + 3 = 0 \quad\Rightarrow\quad (2u - 1)(u - 3) = 0 \quad\Rightarrow\quad u = \tfrac{1}{2} \text{ or } u = 3

Cube both sides to undo $u = x^{\frac{1}{3}}$ : $x = \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ or $x = 3^3 = 27$ .

(Cube roots can be negative, so no value of $u$ would have been rejected here.)

A bracket as the hidden variable

Solve $\left(x^2 - 2x\right)^2 - 11\left(x^2 - 2x\right) + 24 = 0$ .

Solution

Do not expand. Let $u = x^2 - 2x$ :

u^2 - 11u + 24 = 0 \quad\Rightarrow\quad (u - 3)(u - 8) = 0 \quad\Rightarrow\quad u = 3 \text{ or } u = 8

$x^2 - 2x = 3$ : $x^2 - 2x - 3 = 0$ , $(x - 3)(x + 1) = 0$ , so $x = 3$ or $x = -1$ .
$x^2 - 2x = 8$ : $x^2 - 2x - 8 = 0$ , $(x - 4)(x + 2) = 0$ , so $x = 4$ or $x = -2$ .

Solutions: $x = -2, -1, 3, 4$ .

A trigonometric quadratic

Solve $\tan^2 x = 1 + \tan x$ for $0^\circ \le x \le 180^\circ$ .

Solution

Rearrange and let $u = \tan x$ :

u^2 - u - 1 = 0 \quad\Rightarrow\quad u = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}

So $\tan x = 1.618\ldots$ or $\tan x = -0.618\ldots$ . Any real value is possible for $\tan$ , so neither is rejected.

$\tan x = 1.618\ldots$ : $x = \tan^{-1}(1.618\ldots) = 58.3^\circ$ .
$\tan x = -0.618\ldots$ : the calculator gives $-31.7^\circ$ , outside the interval. Since $\tan$ repeats every $180^\circ$ , $x = -31.7^\circ + 180^\circ = 148.3^\circ$ .

Solutions: $x = 58.3^\circ$ and $x = 148.3^\circ$ (to 1 decimal place).

Solving trigonometric equations in an interval is covered fully in Solving trigonometric equations.

Reasoning about the number of roots

Show that, for every positive constant $k$ , the equation $x^4 + 2x^2 - k = 0$ has exactly two real roots.

Solution

Let $u = x^2$ : $u^2 + 2u - k = 0$ , so

u = \frac{-2 \pm \sqrt{4 + 4k}}{2} = -1 \pm \sqrt{1 + k}

Since $k > 0$ , $\sqrt{1 + k} > 1$ . So

$u = -1 + \sqrt{1 + k} > 0$ , which gives two real values $x = \pm\sqrt{-1 + \sqrt{1 + k}}$ ;
$u = -1 - \sqrt{1 + k} < 0$ , which is impossible for $u = x^2$ .

Hence there are exactly two real roots.

Watch out

Stopping at $u$ . The question asks for $x$ . Always back-substitute.

Losing the negative root. $x^2 = 4$ has two solutions, $x = 2$ and $x = -2$ .

Keeping impossible values. Reject $x^2 < 0$ , $\sqrt{x} < 0$ , and $\sin x$ or $\cos x$ outside $[-1, 1]$ , and say why.

Expanding when you should substitute. Expanding $\left(x^2 - 2x\right)^2$ gives a quartic that is much harder to factorise.

Squaring both sides instead. Rearranging $6x - 2 = -\sqrt{x}$ and squaring can work, but it introduces false roots that must be checked. The substitution method avoids this.

Exam tip

Questions often come in two parts: "(a) Solve $2y^2 - 5y + 2 = 0$ . (b) Hence solve $2x^4 - 5x^2 + 2 = 0$ ." Use part (a) directly in (b); do not start again.
Write the substitution explicitly ("let $u = x^2$ "). It makes your working clear and earns the method mark even if a later slip occurs.
Give exact answers (surds, fractions) unless the question says otherwise. For trigonometric versions, give angles to 1 decimal place in degrees or 3 significant figures in radians.
"Find all the solutions" or "solve completely" signals that there are several, often four. Count them before you move on.

Summary

Look for two powers where one is double the other, such as $x^4$ and $x^2$ , $x$ and $\sqrt{x}$ , $\tan^2 x$ and $\tan x$ .
Substitute $u$ for the hidden variable, solve the quadratic in $u$ , then back-substitute.
Reject impossible values with a reason: squares and square roots are not negative, sine and cosine lie in $[-1, 1]$ .
$x^2 = a$ with $a > 0$ gives two solutions, $\pm\sqrt{a}$ .
If the hidden variable is a whole bracket, do not expand it.

Practice questions

Question

Solve $x^4 - 10x^2 + 9 = 0$ .
Solve $x^4 + x^2 - 12 = 0$ .
Solve $x - 5\sqrt{x} + 6 = 0$ .
Solve $2x + 3\sqrt{x} - 2 = 0$ .
Solve $x^6 - 7x^3 - 8 = 0$ .
Solve $\dfrac{6}{x^2} - \dfrac{1}{x} - 1 = 0$ .
(a) Solve $2y^2 - 5y + 2 = 0$ . (b) Hence solve $2x^4 - 5x^2 + 2 = 0$ , giving exact answers.
Solve $2\sin^2 x + \sin x - 1 = 0$ for $0^\circ \le x \le 360^\circ$ .
Solve $\left(x^2 + x\right)^2 - 8\left(x^2 + x\right) + 12 = 0$ .
Find the set of values of the constant $k$ for which the equation $x^4 - 6x^2 + k = 0$ has four distinct real roots.

Answers

$u = x^2$ : $(u - 1)(u - 9) = 0$ , so $x^2 = 1$ or $9$ , giving $x = \pm 1, \pm 3$ .
$u = x^2$ : $(u + 4)(u - 3) = 0$ . $x^2 = -4$ has no real solutions; $x^2 = 3$ gives $x = \pm\sqrt{3}$ .
$u = \sqrt{x}$ : $u^2 - 5u + 6 = 0$ , $u = 2$ or $3$ , so $x = 4$ or $x = 9$ .
$u = \sqrt{x}$ : $2u^2 + 3u - 2 = 0$ , $(2u - 1)(u + 2) = 0$ . $u = -2$ is impossible, so $\sqrt{x} = \tfrac{1}{2}$ and $x = \tfrac{1}{4}$ .
$u = x^3$ : $(u - 8)(u + 1) = 0$ , so $x^3 = 8$ or $-1$ , giving $x = 2$ or $x = -1$ .
$u = \dfrac{1}{x}$ : $6u^2 - u - 1 = 0$ , $(3u + 1)(2u - 1) = 0$ , so $u = \tfrac{1}{2}$ or $-\tfrac{1}{3}$ , giving $x = 2$ or $x = -3$ .
(a) $(2y - 1)(y - 2) = 0$ , so $y = \tfrac{1}{2}$ or $2$ . (b) $x^2 = \tfrac{1}{2}$ or $x^2 = 2$ , so $x = \pm\dfrac{1}{\sqrt{2}} = \pm\dfrac{\sqrt{2}}{2}$ or $x = \pm\sqrt{2}$ .
$u = \sin x$ : $(2u - 1)(u + 1) = 0$ , so $\sin x = \tfrac{1}{2}$ or $\sin x = -1$ . $\sin x = \tfrac{1}{2}$ : $x = 30^\circ, 150^\circ$ . $\sin x = -1$ : $x = 270^\circ$ . Solutions $30^\circ, 150^\circ, 270^\circ$ .
$u = x^2 + x$ : $(u - 2)(u - 6) = 0$ . $x^2 + x - 2 = 0$ gives $x = 1, -2$ ; $x^2 + x - 6 = 0$ gives $x = 2, -3$ . Solutions $x = -3, -2, 1, 2$ .
$u = x^2$ : $u^2 - 6u + k = 0$ . Four distinct real $x$ need two distinct positive values of $u$ (each gives $\pm\sqrt{u}$ ). The roots are $u = 3 \pm \sqrt{9 - k}$ . Distinct roots need $9 - k > 0$ , so $k < 9$ . The larger root $3 + \sqrt{9 - k}$ is always positive; the smaller, $3 - \sqrt{9 - k}$ , is positive when $\sqrt{9 - k} < 3$ , i.e. $9 - k < 9$ , i.e. $k > 0$ . (If $k = 0$ the smaller root is $u = 0$ , giving only $x = 0$ .) Answer: $0 < k < 9$ .

Spotting the disguise

Practice questions

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