Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.
 A: Let $\sqrt[4]{x}=a$ and $\sqrt[4]{x+1}=b$.
Thus, $a\geq0,$ $b\geq1$, $b^4-a^4=1$ and $$a+b=\sqrt[4]{a^4+b^4}$$ or
$$(a+b)^4=a^4+b^4$$ or $$2ab(2a^2+3ab+2b^2)=0,$$ which gives $$ab=0.$$
Can you end it now.
A: If you raise both sides to the 4th power you get
$$x + \mbox{junk} + x+1 = 2x+1$$
or
$$\mbox{junk} =0.$$
If $x>0$ then junk is positive, and you don't have a solution.
So the only solution is $x=0.$
A: You have $x^{1/4}+(x+1)^{1/4}=(2x+1)^{1/4}$
Raise both sides to the power $4$ and you have:
$$x+(x+1)+4x^{3/4}(x+1)^{1/4}+6x^{1/4}(x+1)^{1/4}+4x^{1/4}(x+1)^{3/4}=2x+1$$
$$x^{1/4}(x+1)^{1/4}\Big(2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}\Big)=0$$
From here we have $x=0$ or $x=-1$ as solution.
Now if $x\neq 0$ and $x\neq -1$, then $2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}=0$
Observe that $2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}=2(x^{1/2}+2x^{1/4}(x+1)^{1/4}+(x+1)^{1/2})-x^{1/4}(x+1)^{1/4}=2\Big(x^{1/4}+(x+1)^{1/4}\Big)^2-x^{1/4}(x+1)^{1/4}$
$$2\Big(x^{1/4}+(x+1)^{1/4}\Big)^2=x^{1/4}(x+1)^{1/4}$$
$$2(2x+1)^{1/2}=x^{1/4}(x+1)^{1/4}$$
Raising both the sides to the power of $4$ gives:
$$16(2x+1)^2=x(x+1)$$
This in fact will give us an quadratic equation. I hope you can solve from here.
A: It is trivial that for any $a>0$ and $b>0$ (just raise both side to the $n^{th}$ power)
$$a^\frac1n +b^\frac1n >(a+b)^\frac1n$$
therefore $\forall x>0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}> \sqrt[4]{2x+1}$$
and for $x=0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}= \sqrt[4]{2x+1}=1$$
