Magpie Tuition · Higher Maths

Quartic equations in $x^2$

Short Rescue Pack · allow 20–25 minutes. Solve over the real numbers throughout. No calculator needed. Before starting: use the zero-product rule and read roots from factors (factors and roots).

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Overview

Big Picture

An equation such as

x^4-10x^2+9=0

looks unfamiliar, but $x^4=(x^2)^2$ . It is a quadratic in $x^2$ . Naming $x^2$ as $u$ makes the familiar structure visible:

u=x^2\quad\Rightarrow\quad u^2-10u+9=0.

The substitution is temporary. Solving for $u$ is only the middle of the calculation: the question asks for $x$ .

Method

Step by Step

Put the equation in the form $ax^4+bx^2+c=0$ , with $a\ne0$ . The middle or constant term may be zero. Terms such as $x^3$ or $x$ do not fit this substitution alone.
Write $u=x^2$ , so $x^4=u^2$ .
Solve $au^2+bu+c=0$ . For factorisable examples, factorise and use the zero-product rule.
For every $u$ -value, return to $x^2=u$ .
List all distinct real $x$ -values. Check them in the original equation.

A short factorising reminder

For $u^2+pu+q$ , look for two numbers with sum $p$ and product $q$ . For example, $-1$ and $-9$ sum to $-10$ and multiply to $9$ , giving $(u-1)(u-9)$ . Check by expanding. For a leading coefficient other than 1, check the first terms, last terms and the two cross-products; example 2 shows this explicitly.

Key idea

The Why

The substitution does not change the original problem when you also retain $u=x^2$ . A real square cannot be negative, so the return step has three cases:

Value found for $u$	What $x^2=u$ gives
$u>0$	Two real values: $x=\sqrt u$ or $x=-\sqrt u$
$u=0$	One value: $x=0$
$u<0$	No real value of $x$ from this branch

$\sqrt9=3$ denotes the non-negative square root. But solving $x^2=9$ gives both $3$ and $-3$ , since both square to 9.

Worked example

Worked examples

1. Four real solutions

Solve $x^4-10x^2+9=0$ .

\begin{aligned} u&=x^2,\\ u^2-10u+9&=0,\\ (u-1)(u-9)&=0,\\ u&=1\quad\text{or}\quad u=9. \end{aligned}

Return to $x$ in a separate column:

Quadratic result	Return to $x^2$	Values of $x$
$u=1$	$x^2=1$	$-1,1$
$u=9$	$x^2=9$	$-3,3$

$\boxed{x=-3,-1,1,3}$ . For either sign of 1, the original gives $1-10+9=0$ ; for either sign of 3, it gives $81-90+9=0$ .

2. Leading coefficient and exact square roots

Solve $2x^4-7x^2+3=0$ .

Let $u=x^2$ . Then

2u^2-7u+3=(2u-1)(u-3)=0.

Check: $2u^2-6u-u+3=2u^2-7u+3$ . Thus $2u-1=0$ or $u-3=0$ , giving $u=1/2$ or $u=3$ .

x^2=\frac12\Rightarrow x=\pm\sqrt{\frac12}=\pm\frac{\sqrt2}{2}, \qquad x^2=3\Rightarrow x=\pm\sqrt3.

So $\boxed{x=\pm\sqrt2/2,\ \pm\sqrt3}$ . Keep these exact. Squaring the first pair gives $1/2$ : $2(1/4)-7/2+3=0$ . Squaring the second pair gives $3$ : $18-21+3=0$ .

3. One branch has no real solution

Solve $x^4+x^2-6=0$ .

u=x^2:\quad u^2+u-6=(u+3)(u-2)=0.

Hence $u=-3$ or $u=2$ . The equation $x^2=-3$ has no real solutions because a real square is non-negative. From $x^2=2$ , $x=\pm\sqrt2$ .

$\boxed{x=-\sqrt2,\sqrt2}$ . Each gives $4+2-6=0$ . Reject the negative $u$ branch, not negative values of $x$ .

4. A repeated factor does not give four different answers

Solve $x^4-8x^2+16=0$ .

u=x^2:\quad u^2-8u+16=(u-4)^2=0.

The only $u$ -value is 4. Thus $x^2=4\Rightarrow\boxed{x=-2\text{ or }2}$ . Both give $16-32+16=0$ . Do not list the same solution twice simply because a factor repeats.

Common mistake

The Trap

If you reach $u=1$ or $9$ in example 1:

Writing $x=1,9$ may indicate that you have stopped before substituting back.
Writing $x=1,3$ may indicate that the negative square roots were omitted.
Writing $x=-1,-9$ may indicate that you read the constants in the factors as $x$ -values.

The remedy in each case is the same: write $x^2=u$ for each branch, then solve it. A wrong answer alone does not establish its cause.

Practice

Your Turn

Practice only — these are not your fresh-check questions. Solve over the real numbers. Show the substitution, the quadratic factorisation and the return to $x$ .

$x^4-20x^2+64=0$
$x^4-7x^2+10=0$
$3x^4-13x^2+4=0$
$x^4+5x^2-6=0$
$x^4-9x^2=0$
$x^4+5x^2+4=0$

Answers

Fully worked solutions

Show the answers

1

Let $u=x^2$ . Then $u^2-20u+64=(u-4)(u-16)=0$ , since $-4-16=-20$ and $(-4)(-16)=64$ .

$u=4$ or $16$ . Thus $x^2=4\Rightarrow x=\pm2$ ; $x^2=16\Rightarrow x=\pm4$ .

$\boxed{x=-4,-2,2,4}$ . Checks by squared value: $16-80+64=0$ , $256-320+64=0$ .

2

Let $u=x^2$ . Then $u^2-7u+10=(u-2)(u-5)=0$ : the numbers $-2,-5$ have sum $-7$ and product 10.

$x^2=2$ or $5$ , so $\boxed{x=\pm\sqrt2,\ \pm\sqrt5}$ . Checks: $4-14+10=0$ , $25-35+10=0$ .

3

Let $u=x^2$ . Then $3u^2-13u+4=(3u-1)(u-4)=0$ . Expanding gives $3u^2-12u-u+4$ , confirming the middle coefficient.

$3u-1=0\Rightarrow u=1/3$ , or $u-4=0\Rightarrow u=4$ . Thus

x=\pm\sqrt{1/3}=\pm\frac{\sqrt3}{3}\quad\text{or}\quad x=\pm2.

$\boxed{x=\pm\sqrt3/3,\ \pm2}$ . Checks: $3(1/9)-13/3+4=0$ , $48-52+4=0$ .

4

Let $u=x^2$ . Then $u^2+5u-6=(u+6)(u-1)=0$ . Hence $u=-6$ or $1$ .

$x^2=-6$ has no real solution. $x^2=1$ gives $\boxed{x=-1\text{ or }1}$ . Check: $1+5-6=0$ .

5

Let $u=x^2$ . Then $u^2-9u=u(u-9)=0$ . Therefore $u=0$ or $9$ .

$x^2=0\Rightarrow x=0$ ; $x^2=9\Rightarrow x=\pm3$ . So $\boxed{x=-3,0,3}$ . Checks: $0-0=0$ , $81-81=0$ . Dividing the original equation by $x^2$ would lose the zero solution.

6

Let $u=x^2$ . Then $u^2+5u+4=(u+1)(u+4)=0$ , so $u=-1$ or $-4$ .

Neither $x^2=-1$ nor $x^2=-4$ has a real solution. There are no real solutions. As a second check, $x^4\ge0$ , $5x^2\ge0$ and the constant is 4; their sum cannot be zero for real $x$ .

Before moving on: explain why this pack contains equations with zero, two, three or four distinct real solutions. Practising the method is not proof of mastery; a later fresh check is needed.

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