If $a^2 +\frac{2}{a^2}= 16$, then find the value of $\frac{72a^2}{a^4+2+8a^2}.$
Answer & explanation
Correct answer: option 1
If $a^2 +\frac{2}{a^2}= 16$,
then find the value of $\frac{72a^2}{a^4+2+8a^2}$
We can write $\frac{72a^2}{a^4+2+8a^2}$ by dividing numerator and denominator by a2 as = $\frac{72}{a^2+\frac{2}{a^2}+8}$
Put the given value in the equation given above,
= $\frac{72}{16+8}$ = $\frac{72}{24}$ = 3