Equations Reducible to Quadratics

Introduction

Some equations do not look like quadratics at first glance, yet they share the same underlying structure: a term squared, a term to the first power, and a constant. Recognising this hidden quadratic pattern is a key skill tested in CAIE 9709, appearing in Pure Mathematics questions on algebra and trigonometry alike. The technique is simple — introduce a substitution, solve a familiar quadratic, then translate back — but errors in the final "back-substitution" step are a frequent source of lost marks.

Core Concept

An equation is quadratic in some function of $x$ if it can be written in the form

a[f(x)]^2 + b[f(x)] + c = 0

where $f(x)$ is any expression involving $x$ (e.g. $x^2$ , $\tan x$ , $e^x$ , $\sqrt{x}$ ).

The method is:

Spot the pattern — identify the function $f(x)$ that is being "squared" and also appears to the first power.
Substitute $u = f(x)$ to obtain a standard quadratic $au^2 + bu + c = 0$ .
Solve the quadratic in $u$ using any appropriate method (factorisation, quadratic formula, completing the square).
Back-substitute — for each value of $u$ , solve $f(x) = u$ for $x$ .
Check validity — some values of $u$ may yield no real solutions for $x$ , or may fall outside a given domain.

Key Formulae & Definitions

Standard form of a quadratic in $u$ :

au^2 + bu + c = 0, \qquad a \neq 0

Substitution rule:

Original equation	Substitution	Resulting quadratic
$x^4 - 5x^2 + 4 = 0$	$u = x^2$	$u^2 - 5u + 4 = 0$
$x - 3\sqrt{x} + 2 = 0$	$u = \sqrt{x}$	$u^2 - 3u + 2 = 0$
$\tan^2 x = 1 + \tan x$	$u = \tan x$	$u^2 - u - 1 = 0$
$x^{-2} - 3x^{-1} + 2 = 0$	$u = x^{-1}$	$u^2 - 3u + 2 = 0$

Discriminant (to check whether real solutions in $u$ exist):

\Delta = b^2 - 4ac

Condition	Nature of roots
$\Delta > 0$	Two distinct real roots
$\Delta = 0$	One repeated root
$\Delta < 0$	No real roots

Worked Examples

Example 1 — Polynomial equation: $x^4 - 5x^2 + 4 = 0$

Step 1: Identify the pattern. The terms $x^4$ and $x^2$ differ by a square relationship: $x^4 = (x^2)^2$ . So the equation is quadratic in $x^2$ .

Step 2: Substitute $u = x^2$ .

u^2 - 5u + 4 = 0

Step 3: Factorise.

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

u = 1 \quad \text{or} \quad u = 4

Step 4: Back-substitute $u = x^2$ .

For $u = 1$ :

x^2 = 1 \implies x = \pm 1

For $u = 4$ :

x^2 = 4 \implies x = \pm 2

Step 5: State all solutions.

\boxed{x = -2,\ -1,\ 1,\ 2}

Example 2 — Trigonometric equation: $\tan^2 x = 1 + \tan x$ , for $0° \leq x \leq 360°$

Step 1: Rearrange to bring all terms to one side.

\tan^2 x - \tan x - 1 = 0

Step 2: Identify the pattern. This is quadratic in $\tan x$ . Substitute $u = \tan x$ :

u^2 - u - 1 = 0

Step 3: Solve using the quadratic formula (does not factorise nicely).

u = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}

u \approx 1.618 \quad \text{or} \quad u \approx -0.618

Step 4: Back-substitute $u = \tan x$ , solving over $0° \leq x \leq 360°$ .

For $\tan x \approx 1.618$ :

x = \arctan(1.618) \approx 58.3° \quad \text{or} \quad x \approx 58.3° + 180° = 238.3°

For $\tan x \approx -0.618$ :

x = \arctan(-0.618) \approx -31.7°

Adjusting into $[0°, 360°]$ : $x \approx 180° - 31.7° = 148.3°$ or $x \approx 360° - 31.7° = 328.3°$

Step 5: State all solutions (to 1 d.p.).

\boxed{x \approx 58.3°,\ 148.3°,\ 238.3°,\ 328.3°}

Example 3 — Equation in $\sqrt{x}$ : $x - 5\sqrt{x} + 6 = 0$

Step 1: Substitute $u = \sqrt{x}$ (valid for $x \geq 0$ ), so $x = u^2$ .

u^2 - 5u + 6 = 0

Step 2: Factorise.

(u - 2)(u - 3) = 0

u = 2 \quad \text{or} \quad u = 3

Step 3: Back-substitute $u = \sqrt{x}$ .

\sqrt{x} = 2 \implies x = 4

\sqrt{x} = 3 \implies x = 9

Step 4: State all solutions.

\boxed{x = 4 \quad \text{or} \quad x = 9}

Common Mistakes & Examiner Pitfalls

Forgetting negative roots after back-substitution. In Example 1, $x^2 = 4$ gives $x = \pm 2$ , not just $x = 2$ . Examiners will penalise incomplete solution sets.
Ignoring domain restrictions. If $u = \sqrt{x}$ , then $u \geq 0$ . Any solution $u < 0$ from the quadratic must be discarded. Similarly, if $u = \sin x$ and you find $u = 1.5$ , there is no solution since $|\sin x| \leq 1$ .
Not rearranging before substituting. In Example 2, $\tan^2 x = 1 + \tan x$ must be rearranged to $\tan^2 x - \tan x - 1 = 0$ before substituting. Applying the substitution to an unbalanced equation is a common slip.
Substituting back incorrectly. After finding $u = 4$ where $u = x^2$ , some students write $x = 4$ instead of $x = \pm 2$ . Always ask: "What does $f(x) = u$ actually mean, and how many $x$ values does it imply?"
Losing solutions in the trigonometric step. When solving $\tan x = k$ over a given interval, remember that $\tan$ has period $180°$ . Always use the full interval systematically.
Calling the substitution variable $x$ . Using $x$ for both the substitution and the original variable causes confusion. Always use a distinct letter such as $u$ , $t$ , or $y$ .

Practice Questions

Q1. Solve $x^4 - 13x^2 + 36 = 0$ .

<details><summary>Show answer</summary>

Let $u = x^2$ :

u^2 - 13u + 36 = 0

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

u = 4 \quad \text{or} \quad u = 9

Back-substitute:

x^2 = 4 \implies x = \pm 2

x^2 = 9 \implies x = \pm 3

Solutions: $x = -3,\ -2,\ 2,\ 3$

</details>

Q2. Solve $2x - 7\sqrt{x} + 3 = 0$ , where $x \geq 0$ .

<details><summary>Show answer</summary>

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

2u^2 - 7u + 3 = 0

(2u - 1)(u - 3) = 0

u = \tfrac{1}{2} \quad \text{or} \quad u = 3

Back-substitute $u = \sqrt{x}$ :

\sqrt{x} = \tfrac{1}{2} \implies x = \tfrac{1}{4}

\sqrt{x} = 3 \implies x = 9

Solutions: $x = \tfrac{1}{4}$ or $x = 9$

</details>

Q3. Solve $2\cos^2\theta - 3\cos\theta + 1 = 0$ for $0° \leq \theta \leq 360°$ .

<details><summary>Show answer</summary>

Let $u = \cos\theta$ :

2u^2 - 3u + 1 = 0

(2u - 1)(u - 1) = 0

u = \tfrac{1}{2} \quad \text{or} \quad u = 1

Back-substitute $u = \cos\theta$ :

For $\cos\theta = \tfrac{1}{2}$ : $\theta = 60°$ or $\theta = 300°$

For $\cos\theta = 1$ : $\theta = 0°$ or $\theta = 360°$

Solutions: $\theta = 0°,\ 60°,\ 300°,\ 360°$

</details>

Q4. Solve $x^{-2} - 2x^{-1} - 3 = 0$ .

<details><summary>Show answer</summary>

Let $u = x^{-1} = \dfrac{1}{x}$ :

u^2 - 2u - 3 = 0

(u - 3)(u + 1) = 0

u = 3 \quad \text{or} \quad u = -1

Back-substitute $u = \dfrac{1}{x}$ :

\frac{1}{x} = 3 \implies x = \frac{1}{3}

\frac{1}{x} = -1 \implies x = -1

Solutions: $x = \tfrac{1}{3}$ or $x = -1$

</details>

Q5. Find the values of $x$ in the interval $0 \leq x \leq 2\pi$ satisfying $\sin^2 x + \sin x - 2 = 0$ .

<details><summary>Show answer</summary>

Let $u = \sin x$ :

u^2 + u - 2 = 0

(u + 2)(u - 1) = 0

u = -2 \quad \text{or} \quad u = 1

Check validity ( $-1 \leq \sin x \leq 1$ ): $u = -2$ is impossible, discard it.

For $\sin x = 1$ :

x = \frac{\pi}{2}

Solution: $x = \dfrac{\pi}{2}$ (the only solution in $[0, 2\pi]$ )

</details>

Connections

Prerequisites — assumed known:

Solving Quadratic Equations (factorisation, quadratic formula, completing the square) — the core technique applied after substitution.
Quadratic Inequalities — the same substitution strategy can extend to inequalities of the form $[f(x)]^2 + b[f(x)] + c > 0$ .

Leads directly into:

Trigonometric Equations (Pure Mathematics 2/3) — equations such as $2\sin^2 x - \sin x - 1 = 0$ over extended or general solution domains rely entirely on this method.
Exponential and Logarithmic Equations — equations like $e^{2x} - 3e^x + 2 = 0$ use the substitution $u = e^x$ , an identical strategy.
Further Curve Sketching — roots found here correspond to $x$ -intercepts of quartic or composite functions, linking to graph transformations.

Introduction

Core Concept

Key Formulae & Definitions

Worked Examples

Example 1 — Polynomial equation: $x^4 - 5x^2 + 4 = 0$

Example 2 — Trigonometric equation: $\tan^2 x = 1 + \tan x$ , for $0° \leq x \leq 360°$

Example 3 — Equation in $\sqrt{x}$ : $x - 5\sqrt{x} + 6 = 0$

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Introduction

Core Concept

Key Formulae & Definitions

Worked Examples

Example 1 — Polynomial equation: x4−5x2+4=0x^4 - 5x^2 + 4 = 0x4−5x2+4=0

Example 2 — Trigonometric equation: tan⁡2x=1+tan⁡x\tan^2 x = 1 + \tan xtan2x=1+tanx, for 0°≤x≤360°0° \leq x \leq 360°0°≤x≤360°

Example 3 — Equation in x\sqrt{x}x​: x−5x+6=0x - 5\sqrt{x} + 6 = 0x−5x​+6=0

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Example 1 — Polynomial equation: $x^4 - 5x^2 + 4 = 0$

Example 2 — Trigonometric equation: $\tan^2 x = 1 + \tan x$ , for $0° \leq x \leq 360°$

Example 3 — Equation in $\sqrt{x}$ : $x - 5\sqrt{x} + 6 = 0$