Question

1. (solve) $3x^4 - 13x^2 + 18 = 2x^2$

Explanation:

Step1: Rearrange to standard quartic form

Subtract $2x^2$ from both sides.
$3x^4 - 13x^2 + 18 - 2x^2 = 0$
$3x^4 - 15x^2 + 18 = 0$

Step2: Simplify the equation

Divide all terms by 3.
$x^4 - 5x^2 + 6 = 0$

Step3: Substitute $u=x^2$

Rewrite in quadratic form.
$u^2 - 5u + 6 = 0$

Step4: Factor the quadratic

Find two factors of 6 that sum to -5.
$(u - 2)(u - 3) = 0$

Step5: Solve for $u$

Set each factor equal to 0.
$u - 2 = 0 \implies u=2$; $u - 3 = 0 \implies u=3$

Step6: Substitute back $u=x^2$

Solve for $x$ by taking square roots.
For $u=2$: $x^2=2 \implies x=\pm\sqrt{2}$
For $u=3$: $x^2=3 \implies x=\pm\sqrt{3}$

Answer:

$x = -\sqrt{3},\ -\sqrt{2},\ \sqrt{2},\ \sqrt{3}$