Question

If $5a+\frac{1}{3a}=5$, then value of $9a^2+\frac{1}{25a^2}$ is:

• A. $\frac{51}{5}$
• B. $\frac{39}{5}$
• C. $\frac{52}{5}$
• D. $\frac{39}{5}$

Answer 1

Answer: (B)

$\frac{39}{5}$

$5a+\frac{1}{3a}=5$
$multiplyby\frac{3}{5}bothsides$
$\frac{3}{5}(5a+\frac{1}{3a})=\frac{3}{5}\times 5$
$\Rightarrow 3a+\frac{1}{5a}=3$
$squaringbothsides$
$9a^2+\frac{1}{25a^2}+\frac{6}{5}=9$
$\Rightarrow 9a^2+\frac{1}{25a^2}=9-\frac{6}{5}$
$9a^2+\frac{1}{25a^2}=\frac{39}{5}$