SSC Selection Post 2022 PYQ — Question 66

The correct answer is option 3 i.e. 321

Here, $2x + \frac{1}{3x} = 12$

Multiply both sides by 3/2

$2x \times \frac{3}{2} + \frac{1}{3x} \times \frac{3}{2} = 12 \times \frac{3}{2} \Rightarrow 3x + \frac{1}{2x} = 18$

Now square both sides,$\left(3x + \frac{1}{2x}\right)^2 = 18^2$

$9x^2 + \frac{1}{4x^2} + 2 \cdot 3x \cdot \frac{1}{2x} = 324$

$\Rightarrow 9x^2 + \frac{1}{4x^2} + 3 = 324$

$\Rightarrow 9x^2 + \frac{1}{4x^2} = 324 - 3$ = 321