# Solution Set of $\sqrt{x+1}+\sqrt{x-1}=1$

If $x$ is real, then find the solution set of $\sqrt{x+1}+\sqrt{x-1}=1$.

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The answer is supposed to be a null set but I don't know why. – Nikita May 23 '14 at 23:09

No solutions. For $\sqrt{x-1}$ to be well-defined, we must have $x \ge 1$, so $\sqrt{x+1} \ge \sqrt 2$. The left side of your equation is bigger than the right whenever both sides are well-defined.

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How do you know that the left side is always bigger than the right? – Nikita May 23 '14 at 23:13
@Nikita: You're adding something at least $\sqrt 2$ to something that can't be negative. The result is at least $\sqrt 2$, which is bigger than $1$. – user2357112 May 23 '14 at 23:14

Multiplying by $\sqrt{x+1}-\sqrt{x-1}$ you get

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

But this is impossible, $\sqrt{x+1}$ and $\sqrt{x-1}$ are positive, therefore, their sum is bigger than their difference.

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Here is another, slightly, different solution;

We have $\sqrt{x+1}=1-\sqrt{x-1}$, square both sides to get $x+1=1-2\sqrt{x-1}+x-1$. This means $\sqrt{x-1}=-\frac{1}{2}$ and this has no real solutions. Thus, the original equation has no real solutions.

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