For which (real) values of the variable $x$ are the following equalities valid:

а) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$,
b) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=1$,
c) $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=2$.

(The square root without a sign always denotes the non-negative square root!)
I. solution: To prepare for answering the three questions, let's examine the expression

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

as a function of \( x \), taking the absolute value of the square root in each square root as per the note.

The inner square root is real if \( 2 x-1 \geq 0 \), i.e., \( x \geq 1 / 2 \). For such \( x \), both (large) square roots yield real numbers because the first one contains the sum of two non-negative numbers, and for the second one, by examining the difference of the square of the minuend and the subtrahend, if \( x \geq 1 / 2 \), then \( x^{2}-(\sqrt{2 x-1})^{2}=x^{2}-2 x+1=(x-1)^{2} \geq 0 \), and since the bases are positive, we have

$$
x \geq \sqrt{2 x-1}
$$

The equality holds for \( x=1 \).

Since the expressions under the two square roots are conjugates of each other, squaring the expression promises to be simpler. Indeed, after simplification,

$$
y^{2}=2 x+2 \sqrt{x^{2}-(2 x-1)}=2 x+2 \sqrt{(x-1)^{2}}=2 x+2|x-1|
$$

Now, for \( x \leq 1 \), \( x-1 \) is negative or zero, so \( |x-1|=-(x-1)=1-x \), hence \( y^{2}=2 x+2(1-x)=2 \), therefore \( y=\sqrt{2} \), a constant.

For \( x>1 \), \( x-1 \) is positive, so \( |x-1|=x-1 \), thus \( y^{2}=2 x+2(x-1)=4 x-2 \), therefore \( y=\sqrt{4 x-2} \).

Accordingly, the equality in (a) holds for \( x \geq 1 / 2 \) and \( x \leq 1 \), i.e., in the closed interval \((1 / 2,1)\); it does not hold elsewhere; the equality in (b) does not hold anywhere, it is an equation with no solution, because the value of \( \sqrt{4 x-2} \) at \( x=1 \) is \( \sqrt{2} \), and it increases as \( x \) increases, so \( y \geq \sqrt{2} \) if \( x \leq 1 / 2 \); finally, the equality in (c) is an equation whose solution is \( x=3 / 2 \), the equality holds only for this value.

Klimó János (Kaposvár, public school IV. grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_82397c136f020705f2e3g-1.jpg?height=517&width=499&top_left_y=1216&top_left_x=802)

II. solution: To simplify the left-hand sides, we use the substitution \( \sqrt{2 x-1}=z \). According to our agreement on the value of the square root, only non-negative values of \( z \) need to be considered. Expressing \( x \) in terms of \( z \), we get \( x=\left(z^{2}+1\right) / 2 \), so (1) with the substitution becomes:

$$
\begin{gathered}
y=\sqrt{\frac{z^{2}+1}{2}+z}+\sqrt{\frac{z^{2}+1}{2}-z}=\frac{1}{\sqrt{2}}\left(\sqrt{z^{2}+1+2 z}+\sqrt{z^{2}+1-2 z}\right)= \\
=\frac{1}{\sqrt{2}}(z+1+|z-1|)
\end{gathered}
$$

(The absolute value symbol in the first term of the parentheses can be immediately omitted because \( z+1 \geq 1>0 \).) From here, similarly to the above,

$$
\begin{array}{ll}
0 \leq z \leq 1 \quad \text { for } & y=\frac{1}{\sqrt{2}}[z+1+(1-z)]=\sqrt{2} \\
z \leq 1 \quad \text { for } & y=\frac{1}{\sqrt{2}}[z+1+(z-1)]=\sqrt{2 z}
\end{array}
$$

(see the graph).

Accordingly, the equality in (a) holds for all \( x \) that are given by the \( z \) values satisfying \( 0 \leq z \leq 1 \); from \( x=\left(z^{2}+1\right) / 2 \), we have \( 0 \leq z^{2} \leq 1 \), so \( 1 \leq z^{2}+1 \leq 2 \) and \( 1 / 2 \leq x \leq 1 \);

the equality in (b) does not hold for any \( z \) and thus for any \( x \), because for \( z \geq 1 \), the value of the left-hand side is: \( y=\sqrt{2} z \geq \sqrt{2}>1 \);

the equality in (c) holds from \( \sqrt{2} z=2 \) with \( z=\sqrt{2} \), and thus \( x=(2+1) / 2=3 / 2 \).

Note. Several solutions similar to the second one did not exclude the \( z<0 \) values, so they arrived at a more pleasant graph from the form

$$
y=\frac{1}{2}(|z+1|+|z-1|)
$$

which also contains the mirror image of the graph on the \( Y \)-axis, and thus graphically explains the constancy of \( y \) in the middle interval in a way known from the 4th grade textbook of high school.