If $5 x+\frac{1}{3 x}=4$, then what is the value of $9 x^2+\frac{1}{25 x^2} ?$

$\frac{114}{25}$

If $5 x+\frac{1}{3 x}=4$,

then what is the value of $9 x^2+\frac{1}{25 x^2} ?$

Multiply$5 x+\frac{1}{3 x}=4$ with \(\frac{3}{5}\) to get the desired form of the equation.

So, 3x + \(\frac{1}{5x}\) = 4 × \(\frac{3}{5}\)

3x + \(\frac{1}{5x}\) = \(\frac{12}{5}\)

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2 × k × $\frac{1}{K}$

$9 x^2+\frac{1}{25 x^2}$ = (\(\frac{12}{5}\))² – 2 × 3x × \(\frac{1}{5x}\)

$9 x^2+\frac{1}{25 x^2}$ = (\(\frac{144}{25}\)) - \(\frac{6}{5}\)

$9 x^2+\frac{1}{25 x^2}$ = \(\frac{144 - 30}{25}\) = $\frac{114}{25}$