Linear and Quadratic Equations and Inequalities

Lessons

Senior 4 · Mathematics

Linear and Quadratic Equations and Inequalities

5 lessons

Lesson 3 of 5

Equations reducible to quadratic form

25 min

A quadratic in disguise

The equation $x^{4} - 13 x^{2} + 36 = 0$ has degree 4, and there is no "quartic formula" you are expected to know. But look again: it only contains $x^{4} = (x^{2})^{2}$ and $x^{2}$ . Write $t = x^{2}$ and it becomes $t^{2} - 13 t + 36 = 0$ , an ordinary quadratic. Many equations hide a quadratic in this way: a repeated block such as $x^{2} + x$ , a square root $\sqrt{x}$ , or the sum $x + \frac{1}{x}$ . The method is always the same: choose a substitution, solve the quadratic, go back to $x$ , and check.

Definition

Bi-quadratic equation: an equation $a x^{4} + b x^{2} + c = 0$ with $a \ne 0$ . The substitution $t = x^{2}$ (so $t \ge 0$ ) turns it into $a t^{2} + b t + c = 0$ .

Each root $t \gt 0$ gives two values $x = \pm\sqrt{t}$ ; $t = 0$ gives $x = 0$ ; a root $t \lt 0$ gives no real $x$ .

Example: $x^{4} - 5 x^{2} + 4 = 0$ gives $t^{2} - 5 t + 4 = 0$ , $t = 1$ or $t = 4$ , so $x \in \{-2, -1, 1, 2\}$ .

Worked example 1: two bi-quadratic equations

Solve (a) $x^{4} - 13 x^{2} + 36 = 0$ (b) $x^{4} + 3 x^{2} - 4 = 0$ .

Solution

With $t = x^{2}$ : $t^{2} - 13 t + 36 = (t - 4) (t - 9) = 0$ , so $t = 4$ or $t = 9$ . Then $x^{2} = 4$ gives $x = \pm 2$ , and $x^{2} = 9$ gives $x = \pm 3$ . Solution set $\{-3, -2, 2, 3\}$ .
$t^{2} + 3 t - 4 = (t - 1) (t + 4) = 0$ , so $t = 1$ or $t = - 4$ . The value $t = - 4$ is rejected, since $x^{2}$ cannot be negative. So $x = \pm 1$ .

The four x-intercepts of $y = x^{4} - 13 x^{2} + 36$ are the four roots of Worked example 1(a). The graph is symmetric about the $y$ -axis because only even powers of $x$ appear.

Nested expressions

When the same expression appears more than once, call it $t$ . In $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0$ the block $x^{2} + x$ appears twice; in $x - 5\sqrt{x} + 6 = 0$ , $x = (\sqrt{x})^2$ , so the block is $\sqrt{x}$ .

Worked example 2: substitute the repeated block

Solve (a) $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0$ (b) $x - 5\sqrt{x} + 6 = 0$ .

Solution

Let $t = x^{2} + x$ : $t^{2} - 8 t + 12 = (t - 2) (t - 6) = 0$ . If $t = 2$ : $x^{2} + x - 2 = (x - 1) (x + 2) = 0$ , so $x = 1$ or $x = - 2$ . If $t = 6$ : $x^{2} + x - 6 = (x - 2) (x + 3) = 0$ , so $x = 2$ or $x = - 3$ . Four roots: $\{-3, -2, 1, 2\}$ .
Let $t = \sqrt{x} \ge 0$ (so $x \ge 0$ ): $t^{2} - 5 t + 6 = 0$ , $t = 2$ or $t = 3$ , both allowed. Then $x = 4$ or $x = 9$ . Check: $4 - 10 + 6 = 0$ ✓, $9 - 15 + 6 = 0$ ✓.

Definition

Irrational equation: an equation in which the unknown appears under a root sign, such as $\sqrt{x + 5} = x - 1$ .

Extraneous root: a value produced by the working (usually by squaring) that does not satisfy the original equation.

Important

Squaring is not reversible. $A = B \Rightarrow A^2 = B^2$ , but $A^{2} = B^{2}$ only gives $A = \pm B$ . So

\sqrt{A} = B \iff \begin{cases} A = B^2 \\ B \ge 0 \end{cases}

(the condition $A \ge 0$ is then automatic, since $A = B^{2}$ ). State the condition first, or check every root at the end.

Worked example 3: one root survives

Solve $\sqrt{x + 5} = x - 1$ .

Solution Condition: $x - 1 \ge 0$ , that is $x \ge 1$ . Squaring:

x + 5 = x^2 - 2x + 1 \iff x^2 - 3x - 4 = 0 \iff (x - 4)(x + 1) = 0 .

So $x = 4$ or $x = - 1$ . Only $x = 4$ satisfies $x \ge 1$ . Check: $\sqrt{9} = 3 = 4 - 1$ ✓. For $x = - 1$ : $\sqrt{4} = 2$ but $- 1 - 1 = - 2$ , so $- 1$ is extraneous: it solves $-\sqrt{x + 5} = x - 1$ instead.

Squaring $\sqrt{x + 5} = x - 1$ also solves $-\sqrt{x + 5} = x - 1$ (dashed). Its meeting point $x = - 1$ is the extraneous root, which the check removes.

Worked example 4: two square roots

Solve $\sqrt{2x + 3} - \sqrt{x + 1} = 1$ .

Solution Domain: $x \ge -1$ . Isolate one root and square:

\sqrt{2x + 3} = 1 + \sqrt{x + 1} \;\Longrightarrow\; 2x + 3 = 1 + 2\sqrt{x + 1} + x + 1 \;\Longrightarrow\; x + 1 = 2\sqrt{x + 1} .

Let $u = \sqrt{x + 1} \ge 0$ : $u^{2} = 2 u$ , so $u = 0$ or $u = 2$ , giving $x = - 1$ or $x = 3$ .

Check: $x = - 1$ : $\sqrt{1} - \sqrt{0} = 1$ ✓. $x = 3$ : $\sqrt{9} - \sqrt{4} = 1$ ✓. Both are solutions. (Dividing $x + 1 = 2\sqrt{x + 1}$ by $\sqrt{x + 1}$ would have lost $x = - 1$ .)

Definition

Reciprocal (symmetric) equation of degree 4: $a x^{4} + b x^{3} + c x^{2} + b x + a = 0$ with $a \ne 0$ ; the coefficients read the same forwards and backwards. If $x$ is a root, so is $\frac{1}{x}$ .

Method: $x = 0$ is not a root, so divide by $x^{2}$ and put $t = x + \frac{1}{x}$ , using $x^2 + \frac{1}{x^2} = t^2 - 2$ :

a\left(x^2 + \frac{1}{x^2}\right) + b\left(x + \frac{1}{x}\right) + c = 0 \iff a(t^2 - 2) + bt + c = 0 .

Worked example 5: a reciprocal quartic

Solve $2 x^{4} - 9 x^{3} + 14 x^{2} - 9 x + 2 = 0$ .

Solution The coefficients $2, - 9, 14, - 9, 2$ are symmetric. Dividing by $x^{2}$ and putting $t = x + \frac{1}{x}$ :

2(t^2 - 2) - 9t + 14 = 0 \iff 2t^2 - 9t + 10 = 0 \iff (t - 2)(2t - 5) = 0 .

$t = 2$ : $x + \frac{1}{x} = 2 \iff x^2 - 2x + 1 = 0$ , so $x = 1$ (double). $t = \frac{5}{2}$ : $2 x^{2} - 5 x + 2 = 0$ , so $x = 2$ or $x = \frac{1}{2}$ . Solution set $\{\frac{1}{2}, 1, 2\}$ ; notice the pair $2$ and $\frac{1}{2}$ .

Check $x = 2$ : $32 - 72 + 56 - 18 + 2 = 0$ ✓.

A reciprocal equation of odd degree, such as $2 x^{3} + 3 x^{2} + 3 x + 2 = 0$ , always has the root $x = - 1$ : it factorises as $(x + 1) (2 x^{2} + x + 2) = 0$ , and since $2 x^{2} + x + 2$ has $\Delta = 1 - 16 \lt 0$ , the only real root is $- 1$ .

Warning

Keep the condition on the new letter. For $t = x^{2}$ or $t = \sqrt{x}$ you need $t \ge 0$ ; for $t = x + \frac{1}{x}$ with real $x$ , you need $|t| \ge 2$ (by the AM–GM inequality). A value such as $t = 1$ in a reciprocal equation gives $x^{2} - x + 1 = 0$ , which has no real root.

Tip

Before expanding anything, scan the equation for a block that repeats, a power that is the square of another power, or coefficients that read the same both ways. Expanding $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12$ gives a quartic that is much harder to factorise than the quadratic in $t$ .

The Organic Chemistry Tutor: Solving equations in quadratic form using substitution (about 10 min). Watch how the substitution is chosen in each example, and how every value of the new letter is taken back to x.

✦

Practice

Check every root you find in the original equation.

Solve $x^{4} - 10 x^{2} + 9 = 0$ .
Solve $4 x^{4} - 3 x^{2} - 1 = 0$ .
Solve $(x^{2} - 2 x)^{2} - 2 (x^{2} - 2 x) - 3 = 0$ .
Solve $2x - 7\sqrt{x} + 3 = 0$ .
Solve $\sqrt{3x + 1} = x - 3$ .
Solve $\sqrt{x + 7} + \sqrt{x} = 7$ .
Solve the reciprocal equation $x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 1 = 0$ .
Solve $6 x^{4} - 35 x^{3} + 62 x^{2} - 35 x + 6 = 0$ .

Show answersHide answers

$t^{2} - 10 t + 9 = 0$ : $t = 1, 9$ , so $x = \pm 1, \pm 3$ .
$4 t^{2} - 3 t - 1 = (4 t + 1) (t - 1) = 0$ : $t = 1$ ( $t = -\frac{1}{4}$ rejected), so $x = \pm 1$ .
$t = x^{2} - 2 x$ : $t^{2} - 2 t - 3 = 0$ , $t = 3$ or $t = - 1$ . $x^{2} - 2 x - 3 = 0$ : $x = 3, - 1$ . $x^{2} - 2 x + 1 = 0$ : $x = 1$ . Solution set $\{-1, 1, 3\}$ .
$t = \sqrt{x}$ : $2 t^{2} - 7 t + 3 = (2 t - 1) (t - 3) = 0$ , $t = \frac{1}{2}$ or $3$ : $x = \frac{1}{4}$ or $x = 9$ .
Condition $x \ge 3$ . $3x + 1 = x^2 - 6x + 9 \Rightarrow x^2 - 9x + 8 = 0$ , $x = 1$ or $8$ . Only $x = 8$ (check: $\sqrt{25} = 5$ ✓); $x = 1$ is extraneous.
$\sqrt{x + 7} = 7 - \sqrt{x} \Rightarrow x + 7 = 49 - 14\sqrt{x} + x \Rightarrow \sqrt{x} = 3$ , so $x = 9$ . Check: $4 + 3 = 7$ ✓.
$(t^2 - 2) - 4t + 5 = 0 \Rightarrow t^2 - 4t + 3 = 0$ , $t = 1$ or $3$ . $t = 1$ gives no real $x$ . $x + \frac{1}{x} = 3$ : $x^{2} - 3 x + 1 = 0$ , $x = \frac{3 \pm \sqrt{5}}{2}$ .
$6(t^2 - 2) - 35t + 62 = 0 \Rightarrow 6t^2 - 35t + 50 = 0$ , $t = \frac{5}{2}$ or $\frac{10}{3}$ . Then $x = 2, \frac{1}{2}$ and $x = 3, \frac{1}{3}$ .

Quick check

Which substitution turns (x² + 3x)² − 2(x² + 3x) − 8 = 0 into a quadratic equation?

Key points

Bi-quadratic: $t = x^2 \ge 0$ ; nested: $t$ = the repeated block; $t = \sqrt{x} \ge 0$ when $x$ and $\sqrt{x}$ appear.
$\sqrt{A} = B \iff A = B^2$ and $B \ge 0$ : squaring can add extraneous roots, so state the condition or check.
Reciprocal quartic: divide by $x^{2}$ , put $t = x + \frac{1}{x}$ and use $x^2 + \frac{1}{x^2} = t^2 - 2$ ; odd degree: $x = - 1$ is a root.

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Solving Equations In Quadratic Form Using Substitution, The Organic Chemistry Tutor, embedded from YouTube.

1A quadratic in disguise

2Nested expressions

A quadratic in disguise

Nested expressions