Solve the following equation:

$$
\sqrt{2 x+1}+\sqrt{x+1}=1 .
$$
The square roots on the left side of our equation are functions of the real value $x$, if

$$
2 x+1 \geq 0 \quad \text { and } \quad x+1 \geq 0, \quad \text { that is, if } \quad x \geq-\frac{1}{2} \quad \text { and } \quad x \geq-1
$$

Both conditions are simultaneously satisfied by

$$
x \geq-\frac{1}{2} \cdots
$$

By squaring both sides of the given equation, we obtain

$$
2 x+1+2 \sqrt{(2 x+1)(x+1)}+x+1=1 \quad \text { or } \quad 2 \sqrt{(2 x+1)(x+1)}=-3 x-1 \ldots
$$

thus

$$
-3 x-1>0, \quad \text { or } \quad x<-\frac{1}{3} \cdots
$$

Summarizing the conditions under (1) and (3),

$$
-\frac{1}{2} \leq x<-\frac{1}{3} \cdots
$$

Now, squaring both sides of (2):

$$
4\left(2 x^{2}+3 x+1\right)=9 x^{2}+6 x+1 \quad \text { or } \quad x^{2}-6 x-3=0 \ldots
$$

The solution to (5) is

$$
x=\frac{6 \pm \sqrt{36+12}}{2}=3 \pm \sqrt{12}=3 \pm 2 \sqrt{3}
$$

Of these two values, only the negative one can satisfy the original equation: $x^{\prime}=3-2 \sqrt{3}$. This value satisfies the condition (4); indeed,

$$
-\frac{1}{2}<3-2 \sqrt{3} \sim-0.464<-\frac{1}{3}
$$

Substituting the value of $x^{\prime}$ into the original equation:

$$
\sqrt{7-4 \sqrt{3}}+\sqrt{4-2 \sqrt{3}}=\sqrt{(2-\sqrt{3})^{2}}+\sqrt{(\sqrt{3}-1)^{2}}=2-\sqrt{3}+\sqrt{3}-1=1
$$

Károly Taksony (Year. ev. g. VI. o. Bp.)

Addendum. The other root of equation (5), i.e., $x^{\prime \prime}=3+2 \sqrt{3}$, satisfies the equation

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

Indeed,

$$
\sqrt{7+4 \sqrt{3}}-\sqrt{4+2 \sqrt{3}}=\sqrt{(2+\sqrt{3})^{2}}-\sqrt{(1+\sqrt{3})^{2}}=2+\sqrt{3}-(1+\sqrt{3})=1
$$

If we square both sides of equation (6), we get

$$
-2 \sqrt{(2 x+1)(x+1)}=-3 x-1
$$

If we square again, we arrive at equation (5).