## Task 10/63

It is

$$
\sqrt{3 \frac{3}{8}}=3 \sqrt{\frac{3}{8}} ; \quad \sqrt{17 \frac{17}{288}}=17 \sqrt{\frac{17}{288}} ; \quad \sqrt{12 \frac{12}{143}}=12 \sqrt{\frac{12}{143}}
$$

Under what conditions for the numbers contained in the expressions do the equations generally hold?

Hint: The notation $\sqrt{3 \frac{3}{8}}$ is to be understood as a mixed number, i.e., $\sqrt{3 \frac{3}{8}}=\sqrt{3+\frac{3}{8}}$.
Obviously, in any case

$$
\sqrt{a+\frac{a}{b}}=a \frac{a}{b}
$$

If we square both sides of this equation, we get $a+\frac{a}{b}=\frac{a^{2}}{b}$ or, since $a \neq 0$ can be assumed,

$$
1+\frac{1}{b}=\frac{a^{2}}{b} \rightarrow b+1=a^{2} \rightarrow b=a^{2}-1
$$

Substituting this value into (1), we get

$$
\sqrt{a+\frac{a}{a^{2}-1}}=\sqrt{\frac{a^{3}-1}{a^{2}-1}+\frac{a}{a^{2}-1}}=\sqrt{\frac{a^{3}}{a^{2}-1}}=a \sqrt{\frac{a}{a^{2}-1}}
$$

The equations are therefore valid in general if the numerator of the fraction matches the integer part of the radicand and the denominator is equal to the square of the numerator minus one (which is indeed the case in the given examples). Of course, $a=1$ must be excluded because $a^{2}-1=0$.