[tex]\left(a+\dfrac{1}{a}\right)^2=a^2+2a\cdot\dfrac{1}{a}+ \left(\dfrac{1}{a} \right)^2=a^2+2+\dfrac{1}{\ a^2} =15+2=17
\\\;\\
\left(a+\dfrac{1}{a}\right)^2= 17 \Longrightarrow \sqrt{\left(a+\dfrac{1}{a}\right)^2}= \sqrt{17} \Longrightarrow \left|a+\dfrac{1}{a}\right| =\sqrt{17}
\\\;\\
\Longrightarrow a+\dfrac{1}{a}= \pm\sqrt{17}
\\\;\\
.
[/tex]
