## Task B-2.6.

For which real numbers $a$ does the equation $\sqrt{x^{2}-100}=a+x$ have a solution in the set of real numbers?
## Solution.

There are two conditions for the solution: $x^{2}-100 \geqslant 0$ and $a+x \geqslant 0$.

By squaring the given equation, we get

$x^{2}-100=a^{2}+2 a x+x^{2}$, which simplifies to $x=\frac{-a^{2}-100}{2 a}, a \neq 0$.

The expression for $x$ obtained must satisfy the given conditions, so we have

$\left(\frac{-a^{2}-100}{2 a}\right)^{2}-100 \geqslant 0$ and $a+\frac{-a^{2}-100}{2 a} \geqslant 0$.

After simplification, from the first condition we get $\frac{\left(a^{2}-100\right)^{2}}{4 a^{2}} \geqslant 0$ which holds for any real number $a \neq 0$.

From the second condition, we get $\frac{a^{2}-100}{2 a} \geqslant 0$ or $a \in[-10,0\rangle \cup[10,+\infty\rangle$.

Finally, the intersection of the solution sets of these two conditions is the set of all real numbers $a$ for which the given equation has a solution, i.e., the set $a \in[-10,0\rangle \cup[10,+\infty\rangle$.

## Note:

For a student who set both conditions but did not check one of them: deduct 2 points for the first and 3 points for the second.

A student could solve the equation without setting the conditions and verify by arriving at the equality $\sqrt{\left(\frac{a^{2}-100}{2 a}\right)^{2}}=\frac{a^{2}-100}{2 a}$, or $\left|\frac{a^{2}-100}{2 a}\right|=\frac{a^{2}-100}{2 a}$, which will be true only if $\frac{a^{2}-100}{2 a} \geqslant 0$, or $a \in[-10,0\rangle \cup[10,+\infty\rangle$. With this verification method and the correct solution for $x$, the student should be awarded full marks.