Question 67

If $$a^2+\frac{2}{a^2}=16$$, then find the value of $$\frac{72a^2}{a^4+2+8a^2}$$

Solution

Given, $$a^2+\frac{2}{a^2}=16$$

$$\frac{72a^2}{a^4+2+8a^2}=\frac{72a^2}{a^2\left(a^2+\frac{2}{a^2}+8\right)}$$

$$=\frac{72}{a^2+\frac{2}{a^2}+8}$$

$$=\frac{72}{16+8}$$

$$=\frac{72}{24}$$

$$=3$$

Hence, the correct answer is Option D

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