Question

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

• A. $\frac{51}{5}$
• B. $\frac{29}{5}$
• C. $\frac{52}{5}$
• D. $\frac{39}{5}$

Answer 1

The correct option is D $\frac{39}{5}$

Multiply both side by $3/5$.

$\left(\frac{3}{5}\right)5a+\left(\frac{3}{5}\right)\left(\frac{1}{3}a\right)=\left(\frac{3}{5}\right)5$

=> $3a+\left(\frac{1}{5}a\right)=3$

Squaring both sides,

$9a^2+\left(\frac{1}{25a^2}\right)+\left(\frac{6}{5}\right)=9$ {using $(a+b{)}^2=a^2+b^2+2ab$)

=> $9a^2=\left(\frac{1}{25a^2}\right)=\frac{39}{5}$