Question

solve the equation.
$4x^4 - 40x^2 + 36 = 0$
the solution is $x = \pm \square$ and $x = \pm \square$.

Explanation:

Step1: Substitute $u=x^2$

Let $u=x^2$, so the equation becomes:
$4u^2 - 40u + 36 = 0$

Step2: Simplify the quadratic equation

Divide all terms by 4:
$u^2 - 10u + 9 = 0$

Step3: Factor the quadratic

Find two numbers that multiply to 9 and add to -10:
$(u-1)(u-9) = 0$

Step4: Solve for $u$

Set each factor equal to 0:
$u-1=0 \implies u=1$; $u-9=0 \implies u=9$

Step5: Substitute back $u=x^2$

For $u=1$: $x^2=1 \implies x=\pm1$
For $u=9$: $x^2=9 \implies x=\pm3$

Answer:

$x = \pm 1$ and $x = \pm 3$