If $5 x-\frac{5}{x}+6=0$, then $x^2+\frac{1}{x^2}$ is:
Answer & explanation
Correct answer: option 1
If $K-\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 + 2
If $5 x-\frac{5}{x}+6=0$
$5[ x-\frac{1}{x}]=-6$
x - \(\frac{1}{x}\) = \(\frac{-6}{5}\)
$x^2+\frac{1}{x^2}$ = (\(\frac{-6}{5}\))2 + 2 = \(\frac{36}{25}\) + 2 = $\frac{86}{25}$