*Problem*

For which real numbers x do the following equations hold:

(a) |
\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2} |

(b) |
\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=1 |

(c) |
\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=2? |

*Solution*

First, let's determine the domain of the possible values of x. One restrictions comes from the presence of 2x-1 under the radicals. This implies that we are only interested in x\ge \frac{1}{2}.

Second, the expression x-\sqrt{2x-1} must not be negative. So we require x \ge \sqrt{2x-1}, or after squaring, x^2 -2x+1\ge 0. But the latter holds for all real x anyway; we are thus left with a single restriction: x\ge \frac{1}{2}.

Now, let's combine all the equations into one

\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=a |

Squaring gives

2x +2\sqrt{x^2-\sqrt{(2x-1)^{2}}}=2x\sqrt{x^2 -2x+1}=a^2, |

or,

2x +2|x-1)| = a^2. |

Assuming x\ge 1 we obtain 4x-2=a^2 which translates into x=1 for a=\sqrt{2}, and x=\frac{3}{2} for a=2, while a=1 leads to x=\frac{3}{4} contrary to the assumption x\ge 1.

Assuming \frac{1}{2}\le x \lt 1 gives 2=a^2 which only holds for a=\sqrt{2}. In this case the whole segment \frac{1}{2}\le x \lt 1 satisfies the first equation.