Solve the equation with radicals: $x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0$

Question

Solve this equation :

$ x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0 $

I've calculated $ \Delta $ and it is something like this:

$ \sqrt[4]{4} + 4 \sqrt[4]{2} + 4 - 4 \sqrt[4]{8} $

I don't know what to do next.

Answer 1

Hint: Let $a=\sqrt[4]{2}$. So $a^3=\sqrt[4]{8}$. Now just use the quadratic formula

Answer 2

Let us denote $a=\sqrt[4]2$. (Maybe this will help to notice how coefficients are related.)

We can rewrite the original equation as \begin{align*} x^2-xa(1+a)+a^3&=0\\ x^2-xa-xa^2+a^3&=0\\ x(x-a)-a^2(x-a)&=0\\ (x-a^2)(x-a)&=0 \end{align*}

This only has solutions $x_1=a$ and $x_2=a^2$.