# 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.

• If $x\ne0$ or $x\ne-1$, you can divide both sides by $\sqrt{x^2+x}$ and you have a brand new equation on which you can practice your equation solving skills.
– MasB
Aug 4, 2020 at 12:34
• Are you defining the radicals for negative $x$? Otherwise you should be careful about your solution $x=-1$. Aug 4, 2020 at 12:37

## 4 Answers

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.

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 $$x=0.$$

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.

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 $$x=0$$

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