Real roots of the equation $\frac{18}{x^4} + \frac{1}{x^2} = 4$

Question

I'm struggling a bit on the best method to find the real roots of the above equation.

I ended up obtaining an equation of:

$4x^4 - x^2 - 18 = 0$.

Is this correct?

From there on, how should I factorize in order to obtain the roots?

Answer 1

Yes, it is correct. Now let $x^2=y$ and obtain a second degree equation for $y$. Once you find $y$, $x=\pm\sqrt y$ if $y\ge0$.

Answer 2

You will get $$18+x^2=4x^4$$ $$4x^4-x^2-18=0$$

then, substitute $x^2=t$ to get:

$$4t^2-t-18=0$$

Answer 3

HINT: we get $$0=4x^4-x^2-18$$ and set $$x^2=t$$ and you will get a quadratic equation

Answer 4

$\frac{1}{x^2}=\frac{-1\pm\sqrt {73}}{36} \Rightarrow$ $\frac 1x=\pm$ $\sqrt{ \frac{-1\pm\sqrt {73}}{36}}\Rightarrow x=\pm\left(\frac{6}{\sqrt{{-1\pm\sqrt {73}}}}\right)$