Simplify the following expression:

$$
\frac{a}{x}+\sqrt{\frac{a^{2}}{x^{2}}+1}, \quad \text { where } \quad x=\frac{a^{2}-b^{2}}{2 b}
$$
I. solution. The expression under the square root, after substituting the given expression for $x$, transforms as follows:

$$
\begin{gathered}
\frac{a^{2}}{x^{2}}+1=\frac{4 a^{2} b^{2}}{\left(a^{2}-b^{2}\right)^{2}}+1=\frac{4 a^{2} b^{2}+a^{4}-2 a^{2} b^{2}+b^{4}}{\left(a^{2}-b^{2}\right)^{2}}= \\
=\frac{\left(a^{2}+b^{2}\right)^{2}}{\left(a^{2}-b^{2}\right)^{2}}=\left(\frac{\left(a^{2}+b^{2}\right)^{2}}{\left(a^{2}-b^{2}\right)^{2}}\right)^{2}
\end{gathered}
$$

Thus, the positive square root to be used in $K$ is

$$
\frac{a^{2}+b^{2}}{a^{2}-b^{2}}, \quad \text { or } \quad-\frac{a^{2}+b^{2}}{a^{2}-b^{2}}
$$

depending on whether $b^{2}a^{2}$ (if $b^{2}=a^{2}$, $K$ is undefined). Therefore, in the first case,

$$
K^{\prime}=\frac{2 a b}{a^{2}-b^{2}}+\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=\frac{(a+b)^{2}}{(a+b)(a-b)}=\frac{a+b}{a-b}
$$

and in the second case,

$$
K^{\prime \prime}=\frac{2 a b}{a^{2}-b^{2}}-\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=-\frac{(a-b)^{2}}{(a+b)(a-b)}=\frac{b-a}{a+b}
$$

Simplification was permissible because if $b^{2}-a^{2} \neq 0$, then neither $a+b$ nor $a-b$ is 0.

It is also noted that in the division by $x$, the restriction $b \neq 0$ was assumed.

Eva Tongori (Székesfehérvár, Teleki B. Gymnasium II. grade)

Remark. Several solvers stated that $K$ is always positive. This statement is correct because the square root is positive, and the term preceding it is smaller in absolute value, so the sum is positive regardless of its sign. However, the solvers justified this by considering the final simplified form and only the case where both $a$ and $b$ are positive, although the letters can represent not only positive numbers.

II. solution. The structure of the given expression reminds us of the following form of the quadratic formula for the roots of a quadratic equation:

$$
x^{2}+p x+q=0 \text { gives } \quad x_{1,2}=-\frac{p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^{2}-q} .
$$

The equation for which $K$ is one of the roots (specifically, the larger root, where the square root of the discriminant is taken with a + sign), denoting the unknown by $z$, is as follows:

$$
z^{2}-\frac{2 a}{x} z-1=0
$$

Substituting the given expression for $x$ and multiplying, we get

$$
z^{2}-\frac{4 a b}{a^{2}-b^{2}} z-1=0
$$

The negative of the coefficient of the second term can be written as:

$$
\frac{4 a b}{a^{2}-b^{2}}=\frac{(a+b)^{2}-(b-a)^{2}}{a^{2}-b^{2}}=\frac{(a+b)^{2}}{a^{2}-b^{2}}+\frac{(b-a)^{2}}{b^{2}-a^{2}}=\frac{a+b}{a-b}+\frac{b-a}{a+b}
$$

The product of the two terms in the transformation is -1, equal to the known term in (1). And since the coefficient of $z^{2}$ in (1) is 1, we can conclude from the known relationships between the roots and coefficients that the roots of (1) are

$$
\frac{a+b}{a-b} \quad \text { and } \quad \frac{b-a}{a+b}
$$

The two roots have opposite signs because their product is -1, so the larger root, the sought value of $K$, is the positive root. This, considering the penultimate form of (2),

$$
\begin{aligned}
a^{2}-b^{2}>0 \quad \text { in the first term: } & z=K^{\prime}=\frac{a+b}{a-b} \\
b^{2}-a^{2}>0 \quad \text { in the second term: } & z=K^{\prime \prime}=\frac{b-a}{a+b}
\end{aligned}
$$