Solve $ (13+x)^{1/4}+(4-x)^{1/4}=3$

Question

I have to find all the solution of the equation $ (13+x)^{1/4}+(4-x)^{1/4}=3$.

I know the real solution is $-12$ and $3$ but I don't know how to find them.

Answer 1

setting $$a=\sqrt[4]{13+x}$$ and $$b=\sqrt[4]{4-x}$$ we get the system $$a^4+b^4=17$$ and $$a+b=3$$ with $$b=3-a$$ we get the equation $$a^4+(3-a)^4=17$$ it is $$2(-2+a)(a-1)(16-3a+a^2)=0$$ solve this equation

Answer 2

Setting

$$a=4-x$$

and

$$b=13+x$$

we have :

$$a^{1/4}+b^{1/4}=3$$

Solving by

$$b=( 3-a^{1/4})^{4}$$

but

$$b=17-a$$,

therefore

$$17-a= (3-a^{1/4})^{4}$$

This equation has as roots $a=1$ and $a=16$, hence $x=3$ and $x=-12$.