For which real numbers does the following equation hold?


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    $\begingroup$ Two warnings. (1.) After you square, you will get a perfect square inside the radical, and so the radical can be simplified. However, remember that $\sqrt{a^2}$ is $|a|$ and not simply $a$. (2.) Check that the solution you obtain is a true solution. Squaring an equation could introduce extraneous (i.e., fake) solutions that must be discarded. $\endgroup$ – Srivatsan Sep 13 '11 at 3:20
  • $\begingroup$ A bit late now, but it would have been better if you had showed us what progress you had made, and where you got stuck, so we could tailor our answers to your needs (and also so it wouldn't leave the impression that you were simply interested in getting someone to do all your work for you). $\endgroup$ – Gerry Myerson Sep 13 '11 at 4:52
  • $\begingroup$ This question does not show any research effort $\endgroup$ – The Chaz 2.0 Sep 13 '11 at 4:59
  • $\begingroup$ For whatever it is worth, the question you have asked is the second problem of the very first IMO in $1959$. (imo-official.org/problems.aspx) $\endgroup$ – Adhvaitha Sep 13 '11 at 5:27

For fun, we avoid squaring, though squaring would get rid of that irrational $\sqrt{2}$ a bit faster.

Let $2x-1=u^2$, where we can choose $u \ge 0$. Then $x=(u^2+1)/2$. Thus $$x+\sqrt{2x-1}=\frac{u^2+2u+1}{2}.$$ Similarly, $$x-\sqrt{2x-1}=\frac{u^2-2u+1}{2}.$$ Even without squaring, squares are plentiful enough!

It follows that $$\sqrt{x+\sqrt{2x-1}} +\sqrt{x-\sqrt{2x-1}}=\frac{|u+1|}{\sqrt{2}}+\frac{|u-1|}{\sqrt{2}}.$$

So the original equation becomes $$|u+1|+|u-1|=2.$$

It is clear that $|u+1|=u+1$. If $u \ge 1$, then $|u-1|=u-1$, and we conclude that $u=1$ and therefore $x=1$.

If $0 \le u <1$, then $|u-1|=1-u$, and we end up with the equation $2=2$, which puts no further constraints on $u$. We end up with solutions $1/2\le x<1$. Thus the solutions to the original equation are all $x$ such that $1/2\le x\le 1$. Lots of solutions! None could possibly be extraneous, we didn't do any squaring.

Comment: We can also see directly that $(1+\sqrt{2x-1})^2=2x+2\sqrt{2x-1}$, and $(1-\sqrt{2x-1})^2=2x-2\sqrt{2x-1}$. So we can think of our original equation as saying that the sum of the distances of $\sqrt{2x-1}$ from $1$ and $-1$ is equal to $2$. This sum of distances is $2$ precisely if $-1\le \sqrt{2x-1} \le 1$. But square roots are non-negative, so our condition becomes $0 \le \sqrt{2x-1} \le 1$, or equivalently $1/2 \le x \le 1$.


If you square both sides, you strip away two radicals on the left side, and you have a middle term. The two terms in which $\sqrt{2x-1}$ cancel, and the expression under the radical in the middle term is a perfect square, so that radical goes away as well.

So it's easier than it looks. Try it.

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    $\begingroup$ @jack After solving the equation as suggested by Michael, remember to check that what you have are indeed true solutions. Sometimes squaring etc. might produce extraneous solutions that must be discarded. $\endgroup$ – Srivatsan Sep 13 '11 at 2:41
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    $\begingroup$ Actually it's trickier than it looks. Formally, all $x$ are solutions. If you allow complex numbers (the question says $x$ is real, but doesn't say that all expressions in the equation are real), and use the principal branch of the square root, then it looks to me like all $x \le 1$ are solutions. If you only allow square roots of nonnegative reals, you must have $1/2 \le x \le 1$. $\endgroup$ – Robert Israel Sep 13 '11 at 4:35

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