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For which real numbers does the following equation hold?

$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}$

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    For whatever it is worth, the question you have asked is the second problem of the very first IMO in $1959$. (http://www.imo-official.org/problems.aspx)2011-09-13

2 Answers 2

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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 $1/2 \le x \le 1$.

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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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    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 $1/2 \le x \le 1$.2011-09-13