If $\left(a+\frac{1}{a}+3\right)^2=16$, where a is a non-zero real number, then find the value of $a^2+\frac{1}{a^2}$.
Answer & explanation
Correct answer: option 2
= a + \(\frac{1}{a}\) = n
Then, $a^2+\frac{1}{a^2}$ = n2 - 2
If $\left(a+\frac{1}{a}+3\right)^2=16$
= a + \(\frac{1}{a}\) + 3 = -4 ( 4 can be positive or negative)
= a + \(\frac{1}{a}\) = -4 -3 = -7
$a^2+\frac{1}{a^2}$ = (-7)2 - 2 = 47