5.1. For real numbers $a$ and $b$, it is known that $a b=5, \frac{1}{a^{2}}+\frac{1}{b^{2}}=0.6$. Write down all possible values of $a+b$.
Answer: $5, -5$ (All answers)

Solution. Transform the given conditions: $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{(a+b)^{2}-2 a b}{a^{2} b^{2}}=\frac{(a+b)^{2}-10}{25}=0.6$. Then $(a+b)^{2}=0.6 \cdot 25+10=25$. Therefore, the desired value is 5 or -5.

This situation is possible because we can form a quadratic equation based on the found sum and the known product of the values $a$ and $b$, for which they are roots: $x^{2}-5 x+5=0$ or $x^{2}+5 x+5=0$, and in both cases, the equation has roots.