Question

If $3x+\frac{2}{x}=7$, then $9x^2-\frac{4}{x^2}=$

Answer 1

Given: $3x+\frac{2}{x}=7$
Squaring on both sides, we get
${(3x+\frac{2}{x})}^2=7^2$
$(3x{)}^2+2\times 3x\times \frac{2}{x}+{\left(\frac{2}{x}\right)}^2=49$
$9x^2+12+\frac{4}{x^2}=49$
$\therefore 9x^2+\frac{4}{x^2}=49-12$
$9x^2+\frac{4}{x^2}=37\dots \left(1\right)$

Now, ${(3x-\frac{2}{x})}^2=(3x{)}^2-2\times 3x\times \frac{2}{x}+{\left(\frac{2}{x}\right)}^2$
$=9x^2-12+\frac{4}{x^2}$
$=9x^2+\frac{4}{x^2}-12$
$=37-12\left(\text{from equation}1\right)$
$\therefore {(3x-\frac{2}{x})}^2=25$
$3x-\frac{2}{x}=\pm 5\dots \left(2\right)$

$\left(i\right)$ When $3x-\frac{2}{x}=5$
Now, $9x^2-\frac{4}{x^2}=(3x+\frac{2}{x})(3x-\frac{2}{x})$
$=\left(7\right)\left(5\right)$
$\therefore 9x^2-\frac{4}{x^2}=35$

$\left(ii\right)$ When $3x-\frac{2}{x}=-5$
Now, $9x^2-\frac{4}{x^2}=(3x+\frac{2}{x})(3x-\frac{2}{x})$
$=\left(7\right)(-5)$
$\therefore 9x^2-\frac{4}{x^2}=-35$
As, $35$ is given in the option.

Hence, $\left(b\right)$ is the correct option.