Mathematics 9709 · Pure Mathematics 1

Hidden quadratics

Some equations do not look quadratic. But a substitution can turn them into one.
Once you spot the disguise, a hard-looking equation becomes one you can already solve.

The one idea, and the recipe

When the same thing appears both squared and on its own, call it $u$ . Then a hidden quadratic appears.

This has an $x^4$ and an $x^2$ , so it looks too hard to factorise:

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

But $x^4 = (x^2)^2$ . So the same thing, $x^2$ , appears squared and on its own. Let $u = x^2$ , and it becomes a normal quadratic:

$u^2 - 5u + 4 = 0$

Every problem in this topic is the same three steps:

Spot & name. Find the thing that appears squared and on its own; let $u$ equal it.
Solve the quadratic in $u$ , the kind you factorised pages ago.
Change back. Swap $u$ for what it stood for and solve for $x$ (or $\theta$ ): the step people forget.

Step 2 is just §1.1: let the calculator give you the $u$ -roots. The real skill here is spotting the disguise and changing back.

Finish this example in full

$u^2-5u+4=(u-1)(u-4)=0$ gives $u=1$ or $u=4$ .

Change back through $x^2=u$ . A square root needs a $\pm$ , because a positive and a negative both square to the same number:

$x^2=1\Rightarrow x=\pm1$

$x^2=4\Rightarrow x=\pm2$

Answer

x=\pm1 \quad\text{or}\quad x=\pm2

The disguise did hide extra solutions. The two $u$ values became four $x$ values.

y = x⁴ − 5x² + 4 crosses zero four times, at ±1 and ±2: each positive u = x² split into ±x.

Method 1: disguised in x⁴ and x²

Same method, with $u=x^2$ . In the exam, the number in front just makes the factorising a bit harder.

Worked example

Solve $4x^4-37x^2+9=0$

Let $u=x^2$ , so $x^4=u^2$ : $4u^2 - 37u + 9 = 0$
Factorise (working below): $(4u-1)(u-9) = 0$ , so $u = \tfrac14$ or $u = 9$ .
Change back with $x^2=u$ ; both positive, so every root survives.

Show the factorising and the change-back

Two numbers multiplying to $4 \times 9 = 36$ and adding to $-37$ : those are $-1$ and $-36$ . Split the middle term and factor in pairs:

$4u^2 - u - 36u + 9 = 0$

$u(4u-1) - 9(4u-1) = 0$

$(4u-1)(u-9) = 0$

Then change back:

$x^2 = \tfrac14 \;\Rightarrow\; x = \pm\tfrac12$

$x^2 = 9 \;\Rightarrow\; x = \pm 3$

Answer

x = \pm\tfrac12 \quad\text{or}\quad x = \pm 3

Common mistake

A negative

u

gives no real $x$ , because

x^2=u

can never be negative. Reject it. Do not take its square root.

Your turn— tap to reveal the worked answer (9709-style)

Quick check. Solve $x^4 - 13x^2 + 36 = 0$ with $u=x^2$ .

Numbers multiplying to $+36$ , adding to $-13$ : $-4$ and $-9$ . So $(u-4)(u-9)=0$ , then $x^2=4$ and $x^2=9$ .

Answer

x=\pm2\quad\text{or}\quad x=\pm3

Method 2: disguised in √x and x

Since $(\sqrt{x})^2 = x$ , an equation with both $\sqrt{x}$ and $x$ hides a quadratic. Let $u=\sqrt{x}$ .

Take $x - 7\sqrt{x} + 10 = 0$ . With $u=\sqrt{x}$ (so $x=u^2$ ):

$u^2 - 7u + 10 = 0 \;\Rightarrow\; (u-2)(u-5)=0$ , so $u=2$ or $u=5$ .
Change back by squaring: $\sqrt{x}=2\Rightarrow x=4$ and $\sqrt{x}=5\Rightarrow x=25$ .

The trap version examiners like

Take $x - 3\sqrt{x} - 4 = 0$ . With $u=\sqrt{x}$ :

$u^2 - 3u - 4 = 0 \;\Rightarrow\; (u-4)(u+1)=0$

so $u=4$ or $u=-1$ . A square root is never negative, so $\sqrt{x}=-1$ is impossible. Reject it. Only $u=4$ is left:

Answer

\sqrt{x}=4 \;\Rightarrow\; x=16

Method 3: disguised in trig

Use $\cos^2\theta = 1 - \sin^2\theta$ to get down to one trig ratio. Then it is a quadratic.

Worked example

Solve $2\cos^2\theta + \sin\theta = 1$ , for $0^\circ \le \theta \le 360^\circ$

Get to one ratio: swap $\cos^2\theta$ for $1-\sin^2\theta$ .
Tidy to a quadratic: $2\sin^2\theta - \sin\theta - 1 = 0$ .
Let $u=\sin\theta$ : $(2u+1)(u-1)=0$ , so $u=-\tfrac12$ or $u=1$ .
Change back to $\sin\theta$ ; read off the angles.

Show the algebra and the angles

Replace $\cos^2\theta$ with $1-\sin^2\theta$ :

$2(1 - \sin^2\theta) + \sin\theta = 1$

Expand ( $2\times1=2$ , $2\times(-\sin^2\theta)=-2\sin^2\theta$ ):

$2 - 2\sin^2\theta + \sin\theta = 1$

Bring the $1$ across and multiply by $-1$ so the squared term is positive:

$2\sin^2\theta - \sin\theta - 1 = 0$

With $u=\sin\theta$ , numbers multiplying to $2\times(-1)=-2$ and adding to $-1$ are $-2$ and $+1$ :

$(2u+1)(u-1) = 0 \;\Rightarrow\; u = -\tfrac12 \;\text{or}\; u = 1$

Both are between $-1$ and $1$ . $\sin\theta = 1$ at the top of the sine wave, so $\theta = 90^\circ$ .

For $\sin\theta = -\tfrac12$ : base angle $30^\circ$ (since $\sin 30^\circ = \tfrac12$ ), and sine is negative in the 3rd and 4th quadrants:

$\theta = 180^\circ+30^\circ = 210^\circ \;\text{or}\; \theta = 360^\circ-30^\circ = 330^\circ$

Answer

\theta = 90^\circ,\ 210^\circ,\ 330^\circ

Common mistake

A sine or cosine always stays between

-1

and

1

. If your

u

is outside that range (say

\sin\theta = 2

), reject it.

Other disguises

The marks are in changing back and in rejecting impossible roots. Never stop at the $u$ values.

Whenever a thing appears next to that same thing squared, call it $u$ :

Higher powers like $x^6$ & $x^3$ , with $u = x^3$ (since $x^6=(x^3)^2$ ). (9709/12 Jun 2023 Q4)
Trig, using $\sin^2\theta+\cos^2\theta=1$ to reach a quadratic in one ratio. (9709/13 Nov 2022 Q1)
Roots, with $u=\sqrt{x}$ . (9709/11 Nov 2020 Q12)
Geometric progressions that end in a quadratic in the ratio $r$ .

When you square back from $x^2=u$ , keep the $\pm$ . Each positive $u$ gives two values of $x$ . And reject any root a real value cannot take:

Substitution	Reject when…	Why
$u=x^2$	$u<0$	a square is never negative
$u=\sqrt{x}$	$u<0$	a square root is never negative
$u=\sin\theta,\ \cos\theta$	$u\notin[-1,1]$	sine and cosine stay in $[-1,1]$

Now you try

A cube root can be negative. So with $u=x^3$ , both roots are kept.

Solve $8x^6+215x^3-27=0$ . (9709/12 Jun 2023 Q4)

Your turn— tap to reveal the worked answer (9709/12 Jun 2023 Q4)

Spot the disguise: $x^6=(x^3)^2$ , so let $u=x^3$ : $8u^2+215u-27=0$ .

Numbers multiplying to $8\times(-27)=-216$ and adding to $215$ are $216$ and $-1$ , giving $(8u-1)(u+27)=0$ .

So $u=\tfrac18$ or $u=-27$ . Change back by cube-rooting ( $x=\sqrt[3]{u}$ ). A cube root can be negative, so you keep both:

Answer

x=\tfrac12\quad\text{or}\quad x=-3

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Hidden quadratics

Some equations do not look quadratic. But a substitution can turn them into one.
Once you spot the disguise, a hard-looking equation becomes one you can already solve.

The one idea, and the recipe

When the same thing appears both squared and on its own, call it $u$ . Then a hidden quadratic appears.

This has an $x^4$ and an $x^2$ , so it looks too hard to factorise:

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

But $x^4 = (x^2)^2$ . So the same thing, $x^2$ , appears squared and on its own. Let $u = x^2$ , and it becomes a normal quadratic:

$u^2 - 5u + 4 = 0$

Every problem in this topic is the same three steps:

Spot & name. Find the thing that appears squared and on its own; let $u$ equal it.
Solve the quadratic in $u$ , the kind you factorised pages ago.
Change back. Swap $u$ for what it stood for and solve for $x$ (or $\theta$ ): the step people forget.

Step 2 is just §1.1: let the calculator give you the $u$ -roots. The real skill here is spotting the disguise and changing back.

Finish this example in full

$u^2-5u+4=(u-1)(u-4)=0$ gives $u=1$ or $u=4$ .

Change back through $x^2=u$ . A square root needs a $\pm$ , because a positive and a negative both square to the same number:

$x^2=1\Rightarrow x=\pm1$

$x^2=4\Rightarrow x=\pm2$

Answer

x=\pm1 \quad\text{or}\quad x=\pm2

The disguise did hide extra solutions. The two $u$ values became four $x$ values.

y = x⁴ − 5x² + 4 crosses zero four times, at ±1 and ±2: each positive u = x² split into ±x.

Method 1: disguised in x⁴ and x²

Same method, with $u=x^2$ . In the exam, the number in front just makes the factorising a bit harder.

Worked example

Solve $4x^4-37x^2+9=0$

Let $u=x^2$ , so $x^4=u^2$ : $4u^2 - 37u + 9 = 0$
Factorise (working below): $(4u-1)(u-9) = 0$ , so $u = \tfrac14$ or $u = 9$ .
Change back with $x^2=u$ ; both positive, so every root survives.

Show the factorising and the change-back

Two numbers multiplying to $4 \times 9 = 36$ and adding to $-37$ : those are $-1$ and $-36$ . Split the middle term and factor in pairs:

$4u^2 - u - 36u + 9 = 0$

$u(4u-1) - 9(4u-1) = 0$

$(4u-1)(u-9) = 0$

Then change back:

$x^2 = \tfrac14 \;\Rightarrow\; x = \pm\tfrac12$

$x^2 = 9 \;\Rightarrow\; x = \pm 3$

Answer

x = \pm\tfrac12 \quad\text{or}\quad x = \pm 3

Common mistake

A negative

u

gives no real $x$ , because

x^2=u

can never be negative. Reject it. Do not take its square root.

Your turn— tap to reveal the worked answer (9709-style)

Quick check. Solve $x^4 - 13x^2 + 36 = 0$ with $u=x^2$ .

Numbers multiplying to $+36$ , adding to $-13$ : $-4$ and $-9$ . So $(u-4)(u-9)=0$ , then $x^2=4$ and $x^2=9$ .

Answer

x=\pm2\quad\text{or}\quad x=\pm3

Method 2: disguised in √x and x

Since $(\sqrt{x})^2 = x$ , an equation with both $\sqrt{x}$ and $x$ hides a quadratic. Let $u=\sqrt{x}$ .

Take $x - 7\sqrt{x} + 10 = 0$ . With $u=\sqrt{x}$ (so $x=u^2$ ):

$u^2 - 7u + 10 = 0 \;\Rightarrow\; (u-2)(u-5)=0$ , so $u=2$ or $u=5$ .
Change back by squaring: $\sqrt{x}=2\Rightarrow x=4$ and $\sqrt{x}=5\Rightarrow x=25$ .

The trap version examiners like

Take $x - 3\sqrt{x} - 4 = 0$ . With $u=\sqrt{x}$ :

$u^2 - 3u - 4 = 0 \;\Rightarrow\; (u-4)(u+1)=0$

so $u=4$ or $u=-1$ . A square root is never negative, so $\sqrt{x}=-1$ is impossible. Reject it. Only $u=4$ is left:

Answer

\sqrt{x}=4 \;\Rightarrow\; x=16

Method 3: disguised in trig

Use $\cos^2\theta = 1 - \sin^2\theta$ to get down to one trig ratio. Then it is a quadratic.

Worked example

Solve $2\cos^2\theta + \sin\theta = 1$ , for $0^\circ \le \theta \le 360^\circ$

Get to one ratio: swap $\cos^2\theta$ for $1-\sin^2\theta$ .
Tidy to a quadratic: $2\sin^2\theta - \sin\theta - 1 = 0$ .
Let $u=\sin\theta$ : $(2u+1)(u-1)=0$ , so $u=-\tfrac12$ or $u=1$ .
Change back to $\sin\theta$ ; read off the angles.

Show the algebra and the angles

Replace $\cos^2\theta$ with $1-\sin^2\theta$ :

$2(1 - \sin^2\theta) + \sin\theta = 1$

Expand ( $2\times1=2$ , $2\times(-\sin^2\theta)=-2\sin^2\theta$ ):

$2 - 2\sin^2\theta + \sin\theta = 1$

Bring the $1$ across and multiply by $-1$ so the squared term is positive:

$2\sin^2\theta - \sin\theta - 1 = 0$

With $u=\sin\theta$ , numbers multiplying to $2\times(-1)=-2$ and adding to $-1$ are $-2$ and $+1$ :

$(2u+1)(u-1) = 0 \;\Rightarrow\; u = -\tfrac12 \;\text{or}\; u = 1$

Both are between $-1$ and $1$ . $\sin\theta = 1$ at the top of the sine wave, so $\theta = 90^\circ$ .

For $\sin\theta = -\tfrac12$ : base angle $30^\circ$ (since $\sin 30^\circ = \tfrac12$ ), and sine is negative in the 3rd and 4th quadrants:

$\theta = 180^\circ+30^\circ = 210^\circ \;\text{or}\; \theta = 360^\circ-30^\circ = 330^\circ$

Answer

\theta = 90^\circ,\ 210^\circ,\ 330^\circ

Common mistake

A sine or cosine always stays between

-1

and

1

. If your

u

is outside that range (say

\sin\theta = 2

), reject it.

Other disguises

The marks are in changing back and in rejecting impossible roots. Never stop at the $u$ values.

Whenever a thing appears next to that same thing squared, call it $u$ :

Higher powers like $x^6$ & $x^3$ , with $u = x^3$ (since $x^6=(x^3)^2$ ). (9709/12 Jun 2023 Q4)
Trig, using $\sin^2\theta+\cos^2\theta=1$ to reach a quadratic in one ratio. (9709/13 Nov 2022 Q1)
Roots, with $u=\sqrt{x}$ . (9709/11 Nov 2020 Q12)
Geometric progressions that end in a quadratic in the ratio $r$ .

When you square back from $x^2=u$ , keep the $\pm$ . Each positive $u$ gives two values of $x$ . And reject any root a real value cannot take:

Substitution	Reject when…	Why
$u=x^2$	$u<0$	a square is never negative
$u=\sqrt{x}$	$u<0$	a square root is never negative
$u=\sin\theta,\ \cos\theta$	$u\notin[-1,1]$	sine and cosine stay in $[-1,1]$

Now you try

A cube root can be negative. So with $u=x^3$ , both roots are kept.

Solve $8x^6+215x^3-27=0$ . (9709/12 Jun 2023 Q4)

Your turn— tap to reveal the worked answer (9709/12 Jun 2023 Q4)

Spot the disguise: $x^6=(x^3)^2$ , so let $u=x^3$ : $8u^2+215u-27=0$ .

Numbers multiplying to $8\times(-27)=-216$ and adding to $215$ are $216$ and $-1$ , giving $(8u-1)(u+27)=0$ .

So $u=\tfrac18$ or $u=-27$ . Change back by cube-rooting ( $x=\sqrt[3]{u}$ ). A cube root can be negative, so you keep both:

Answer

x=\tfrac12\quad\text{or}\quad x=-3