Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$

Question

Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$

This is my textbook contest exercise.

$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$

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

I know that

$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$

So, we will have some cube powers which introduce still $\sqrt [4]{\cdot}$ roots.

This makes more complicated the equation. How can I find the minimal polynomial using a more simple way?

Answer 1

Let $x=u^4,\;2-x=v^4$ with the restriction $u,v≥0$, then we have:

$$\begin{align}&\begin{cases}u-v=1\\u^4+v^4=2\end{cases} \\\implies &v^4+(v+1)^4=2\end{align}$$

Then, using the key substitution $v+\frac 12=y$ leads to:

$$\begin{align}&\left(y-\frac 12\right)^4+\left(y+\frac 12\right)^4=2\\ \implies &16y^4+24y^2-15=0\end{align}$$

Finally, substitute $y^2=z$, you obtain:

$$16z^2+24z-15=0$$

The solution is completed by reversing the steps.

Answer 2

In this answer, I wanted to construct the trigonometric method that requires significantly less computation.

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Since $0<x<2$, making the substitution $x=2\sin^2\theta$, where $0<\theta<\frac {\pi}{2}$, we have:

$$\begin{align}&\sqrt {\sin\theta}-\sqrt {\cos\theta}=\frac {1}{\sqrt [4]{2}}\\ \implies&\sin\theta+\cos\theta=\frac{1}{\sqrt 2}+\sqrt {2\sin (2\theta)}\\ \implies&2\sqrt {\sin (2\theta)}+\sin (2\theta)=\frac 12\end{align}$$

Then the substitution $\sqrt {\sin (2\theta)}=u$, leads to:

$$\begin{align}&2u^2+4u-1=0,~u>0\\ \implies &\sin2\theta=\left(\frac {\sqrt 6-2}{2}\right)^2\\ \implies &\theta =\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\end{align}$$

Thus, the final answer is:

$$\bbox[5px,border:2px solid #C0A000]{x=2\sin^2\left(\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\right)}$$