# (solved)Question 2 SSC-CGL 2019 June 13 Shift 1

If $\displaystyle \sqrt x + \frac{1}{\sqrt x} = \sqrt 6$, then $\displaystyle x^2 + \frac{1}{x^2}$ =?
(Rev. 18-Jun-2024)

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### Question 2SSC-CGL 2019 June 13 Shift 1

If $\displaystyle \sqrt x + \frac{1}{\sqrt x} = \sqrt 6$, then $\displaystyle x^2 + \frac{1}{x^2}$ = ?

1. 18
2. 14
3. 16
4. 12
This is a very regularly asked question for the past 10 years.

### Solution in Brief

A shortcut trick: Square of the power on LHS = square of the value on the right minus 2. So $\displaystyle x + \text{. . . }$ $\displaystyle = (\sqrt 6)^2 - 2 = 4$. Squaring once again, $\displaystyle x^2 + \text{. . . }$ = $\displaystyle 4^2 - 2$ = 14 ans!

### Solution in Detail

Given $\displaystyle \sqrt x + \frac{1}{\sqrt x} = \sqrt 6 \text{. . . (1)}$

Recall that $\displaystyle (a + b)^2 = a^2 + 2ab + b^2$

Square both sides of (1) to get LHS:

$\displaystyle (\sqrt x)^2 + 2\cdot \sqrt x \cdot \frac{1}{\sqrt x} + \bigg(\frac{1}{\sqrt x}\bigg)^2$

and $\displaystyle \therefore x + 2 + \frac 1x = (\sqrt 6)^2$

which gives $\displaystyle x + \frac 1x = 4 \text{ . . . (2)}$

Squaring (2), we get

$\displaystyle x^2 + 2\cdot x^2 \cdot\frac {1}{x^2} + \frac {1}{x^2} = 16$

$\displaystyle \therefore x^2 + \frac{1}{x^2} = 16 - 2 = 14$ Ans!