Solve the following equation:

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

For what values of $m$ does a real solution exist, and how many are there?
Our equation is equivalent to the following system of equations:

$$
\sqrt{x+1}=u, \quad \sqrt{2 x+1}=v, \quad u-v=m
$$

or

$$
x+1=u^{2}, \quad 2 x+1=v^{2}, \quad u-v=m \ldots
$$

where

$$
u>0, \quad v>0
$$

By eliminating $v$:

$$
v=u-m, \quad x+1=u^{2}, \quad 2 x+1=(u-m)^{2} \ldots
$$

or

$$
v=u-m, \quad x=u^{2}-1, \quad f(u) \equiv u^{2}+2 m u-m^{2}-1=0 \ldots
$$

where

$$
u>0 \text { and } u-m>0, \text { i.e., } u>m \ldots
$$

We need to solve equation (3) for $u$, taking into account conditions (4).

I. Let $m<0$. Here, $u>0$ is satisfied, $u>m$. However, if $m<0$, then $u=1$ and $x=u^{2}-1=0$.

II. Let $m=0$. Here, $u>0$ is satisfied, $u=1$ and $x=u^{2}-1=0$.

III. Let $m>0$. Now, it is sufficient to satisfy the condition $u>m$, because in this case $u>0$. Since equation (3) has real roots with opposite signs, only the positive one can meet the condition if it is greater than $m$.

The function $f(u)$ is a quadratic function of $u$ which is negative between its zeros. The positive zero is greater than $m$ if

$$
f(m)<0, \text { i.e., } m^{2}+2 m^{2}-m^{2}-1<0, \text { i.e., } 2 m^{2}-1<0, \text { i.e., } m<\frac{\sqrt{2}}{2}
$$

If $m>\frac{\sqrt{2}}{2}$, the equation has no solution.

Freud Géza (Berzsenyi g. VI. o. Bp. V.)

Note: Let's examine the variation of the function

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

Real values are obtained only if $x \geq-\frac{1}{2}$, i.e., the domain of the function extends from $x=-\frac{1}{2}$ to $x=+\infty$.

If $x=-\frac{1}{2}, y=\frac{\sqrt{2}}{2}$; at $x=0$, $y=0$; if $x=+\infty, y=-\infty$.

The function is continuous and constantly decreasing in the previously mentioned interval because its derivative

$$
\begin{gathered}
y^{\prime}=\frac{1}{2 \sqrt{x+1}}-\frac{1}{\sqrt{2 x+1}}<0, \text { if only } x \geq-\frac{1}{2} \\
x=-\frac{1}{2} \text { at } y^{\prime}=-\infty
\end{gathered}
$$

Let's plot the parts of the following two parabolas above the $X$-axis:
a) $y_{1}=\sqrt{x+1}$,
b) $y_{2}=\sqrt{2 x+1}$.

The vertex of the parabola in (a) is at the point $(x=-1, y=0)$, and the vertex of the parabola in (b) is at $x=-\frac{1}{2}, y=0$. The two parabolas intersect at the point $(x=0, y=1)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2161714111c2bf184b74g-2.jpg?height=393&width=555&top_left_y=427&top_left_x=774)

If we subtract the ordinates of the points corresponding to the parabola in (a) from the ordinates of the corresponding points of the parabola in (b) and measure these differences vertically from the $X$-axis (at the corresponding $x$-position), the endpoints of these differences will form the curve

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

The ordinates of the points on this curve vary from $\frac{\sqrt{2}}{2}$ to $0$, and then from $0$ to $-\infty$.

If we draw a line parallel to the $X$-axis at $y=m$, this line will intersect the curve at exactly one point, provided $0 \leq m \leq \frac{\sqrt{2}}{2}$, or $m<0$.