The function is $$ f(x)=\sqrt{x+\sqrt{x}} $$ I know that you need to set up the equation $$\sqrt{x_1+\sqrt{x_1}}=\sqrt{x_2+\sqrt{x_2}}$$ and you have to solve step by step until you get $x_1=x_2$. But I am having difficulty figuring out what to do. I have tried squaring both sides, factoring, then completing the square but nothing has worked. All answers appreciated. Please keep in mind that I need every step, thank you.


Can you use the fact that the (nonnegative branch of the) square root is strictly increasing?

If so, one way to prove it would be to notice that if for some $0\leq x_1< x_2$ we have




and so


but since we are assuming $x_1<x_2$, the left side of the equation is negative, and the right side is positive.

Just in case, to prove that the square root is strictly increasing, let $x_1,x_2$ be nonnegative real numbers with $x_1<x_2$. If we suppose that $\sqrt{x_1}\geq \sqrt{x_2}$ then, since we are dealing with nonnegative numbers, squaring the inequality gives us $x_1\geq x_2$ which contradicts our hypothesis.

  • $\begingroup$ perfect thank you so much $\endgroup$ – Matt Nov 15 '13 at 3:56
  • $\begingroup$ also how do you type the square roots and other math symbols because my equation was messy. thank you $\endgroup$ – Matt Nov 15 '13 at 3:59
  • $\begingroup$ sorry i was thinking and i realized that i need to simplify it to x1=x2 sorry about notation and my lack of knowledge im only in precalculus, $\endgroup$ – Matt Nov 15 '13 at 4:09
  • $\begingroup$ im confused, i did an example with x1=4 x2=9 4-9= -5 sqrt(9)-sqrt(4)=1 so why does x1-x2=sqrt(x2)-sqrt(x1) prove that x1=x2 $\endgroup$ – Matt Nov 15 '13 at 4:18
  • $\begingroup$ You can find a very nice basic tutorial about inserting math symbols in the link provided by T.Bongers in his comment to your question. Now, the proof that I wrote uses contradiction (see en.wikipedia.org/wiki/Proof_by_contradiction), your example illustrates that if we start with 4 and 9 and we suppose that their values on f are the same, we get that -5=1, which of course is not true. If you want I can write a direct argument, but N.S. basically wrote this in his answer. By the way, if you consider some answer satisfactory maybe you could consider accepting it :-) $\endgroup$ – Aldo Guzmán Sáenz Nov 15 '13 at 5:06

The calculus way: We can compute $f'$ easily enough to find that if $f(x) = \left(x + x^{1/2}\right)^{1/2}$ then

$$f'(x) = \frac{1}{2} \left(x + x^{1/2}\right)^{-1/2} \left(1 + \frac{1}{2} x^{-1/2}\right)$$

This is positive for all $x > 0$, so the function is one-to-one.



If $x_1 <x_2$ then $\sqrt{x_1} < \sqrt{x_2}$. Adding the two inequalities you get $x_1+\sqrt{x_1} < x_2+\sqrt{x_2}$. Finally take the square root:

$$\sqrt{x_1+\sqrt{x_1}} < \sqrt{x_2+\sqrt{x_2}} \,.$$


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