How to solve that?

Question

I have no idea about how to solve the following: $$\sqrt[4]{13x+1} + \sqrt[4]{4x-1} = 3\sqrt[4]{x}$$

Could somebody help me, please?

Answer 1

Let $t^4=4-\frac1x$

$$\sqrt[4]{13x+1} + \sqrt[4]{4x-1} = 3\sqrt[4]{x}\iff \sqrt[4]{17-t^4} + t = 3 \iff 17-t^4=(3-t)^4$$

$$\iff 2 t^4 - 12 t^3 + 54 t^2 - 108 t + 64=0 \iff 2 (t - 1) (t - 2) (t^2 - 3 t + 16)=0$$

thus

• $t=2 \implies 16=4-\frac1x \implies x=-\frac1{12}$ (complex solution)
• $t=1 \implies 1=4-\frac1x \implies x=\frac13$ (real solution)

Answer 2

Let $\sqrt[4]{13x+1}=a$ and $ \sqrt[4]{4x-1}=b$.

Thus, $a\geq0$ and $b\geq0$ and $$a+b=3\sqrt[4]{\frac{a^4+b^4}{17}},$$ which is homogeneous and $$\frac{a^4-1}{13}=\frac{b^4+1}{4}.$$ The first gives after squaring of the both sides: $$(a-2b)(2a-b)(16a^2+23ab+16b^2)=0.$$ Since $16a^2+23ab+16b^2=0$ gives $a=b=0,$ which is impossible,

we get $(a-2b)(2a-b)=0$ only. I think now you can end it.