How do I determine the values of x when f(x) = 48 when f(x) = x^{4} - 5x^{2} + 4. Please and thank you!
Let $\displaystyle w = x^2$, then solve for w, then take square roots to find x values.
Some polynomials are "secretly" solvable as quadratics. Here are some other eamples:
$\displaystyle 4x^{20}-4x^{10}+1 = 0, \ q^8 + 7q^4 +6 = 0, \ z^{14}-z^7-56 = 0$
$\displaystyle x^4 - 5x^2 + 4 = 48$
$\displaystyle x^4 - 5x^2 = 44$
complete the square ...
$\displaystyle x^4 - 5x^2 + \frac{25}{4} = 44 + \frac{25}{4}$
$\displaystyle \left(x^2 - \frac{5}{2}\right)^2 = \frac{201}{4}$
$\displaystyle x^2 - \frac{5}{2} = \pm \frac{\sqrt{201}}{2}$
assuming you're looking for real roots only ... $\displaystyle x^2 > 0$ for all $\displaystyle x \ne 0$
$\displaystyle x^2 = \frac{5 + \sqrt{201}}{2}$
$\displaystyle x = \pm \sqrt{\frac{5 + \sqrt{201}}{2}}$