Question

1. $x^{4}-16x^{2}=x^{2}+18$

Explanation:

Step1: Rearrange to standard quartic form

$x^4 - 16x^2 - x^2 - 18 = 0$
$x^4 - 17x^2 - 18 = 0$

Step2: Substitute $u=x^2$

Let $u = x^2$, so the equation becomes:
$u^2 - 17u - 18 = 0$

Step3: Factor the quadratic

$(u - 18)(u + 1) = 0$

Step4: Solve for $u$

Set each factor equal to 0:
$u - 18 = 0 \implies u = 18$
$u + 1 = 0 \implies u = -1$

Step5: Substitute back $u=x^2$

For $u=18$: $x^2 = 18 \implies x = \pm\sqrt{18} = \pm3\sqrt{2}$
For $u=-1$: $x^2 = -1 \implies x = \pm i$ (complex solutions)

Answer:

Real solutions: $x = 3\sqrt{2}$, $x = -3\sqrt{2}$
Complex solutions: $x = i$, $x = -i$