Question

$a+\dfrac{1}a=\dfrac{5}2.\ Ar\breve{a}ta\c{t}i\ c\breve{a}\ a^2+\dfrac{1}{a^2}=\dfrac{17}4.$ 

Answer 1

Salut,

$\left(a+\dfrac{1}a\right)^2=a^2+2\cdot a\cdot\dfrac{1}a+\dfrac{1}{a^2}=a^2+2+\dfrac{1}{a^2}\Rightarrow\\\\\Rightarrow a^2+\dfrac{1}{a^2}=\left(a+\dfrac{1}a\right)^2-2=\left(\dfrac{5}2\right)^2-2=\dfrac{25}4-2=\dfrac{17}4.$

Green eyes.