If $a - \frac{12}{a} = 1$, where a > 0, then the value of $a^2 + \frac{16}{a^2}$ is:
Answer & explanation
Correct answer: option 2
If $a - \frac{12}{a} = 1$
Put the value of a = 4 that will satisfy the given equation.
then the value of $a^2 + \frac{16}{a^2}$ is
$a^2 + \frac{16}{a^2}$ = $4^2 + \frac{16}{4^2}$ = 16 + 1
$a^2 + \frac{16}{a^2}$ = 17