Quadratics: Question 5

Worked solution

Part (a): Making the substitution

Let $u = x^2$. Then $x^4 = (x^2)^2 = u^2$, so the equation $x^4 - 7x^2 - 18 = 0$ becomes:

$u^2 - 7u - 18 = 0$

as required.

Part (b): Solving the quadratic in $u$

Factorise, looking for two numbers with product $-18$ and sum $-7$: these are $-9$ and $2$.

$u^2 - 7u - 18 = (u - 9)(u + 2) = 0$

$u = 9 \qquad \text{or} \qquad u = -2$

Check using the quadratic formula:

$\text{discriminant} = (-7)^2 - 4(1)(-18) = 49 + 72 = 121$

$u = \frac{7 \pm \sqrt{121}}{2} = \frac{7 \pm 11}{2} = 9 \text{ or } -2$

This agrees exactly with the factorised solution.

Part (c): Returning to $x$, and rejecting an extraneous branch

Recall $u = x^2$. Since $x$ is real, $x^2 \ge 0$ for every real $x$, so $u$ can never be negative.

• $u = -2$ would require $x^2 = -2$, which has no real solutions, so this value of $u$ must be rejected.
• $u = 9$ gives $x^2 = 9$, so $x = \pm 3$.

So the real solutions of $x^4 - 7x^2 - 18 = 0$ are $x = 3$ and $x = -3$.

Check: at $x=3$, $x^4 - 7x^2 - 18 = 81 - 63 - 18 = 0$. ✓ (By symmetry, $x=-3$ gives the same value, since only even powers of $x$ appear.)

Final answers

• (a) $u^2 - 7u - 18 = 0$ (shown)
• (b) $u = \boxed{9}$ or $u = \boxed{-2}$
• (c) $x = \boxed{3}$ or $x = \boxed{-3}$ (rejecting $u=-2$ as no real $x$ satisfies $x^2=-2$)