[2] If x = 3 + 2\sqrt{2}, then the value of \left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right) is:

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Aptitude

TNPSC G4

10th CBSE Maths

If x = 3 + 2$\sqrt{2}$, then the value of $\left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)$ is:

2$\sqrt{2}$

3$\sqrt{3}$

Explanation:

$\left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)^{2}=x+\dfrac{1}{x}-2$

$= (3 + 2\sqrt{2}) +\dfrac{1}{(3 + 2\sqrt{2})}-2$

$= (3 + 2\sqrt{2}) +\dfrac{1}{(3 + 2\sqrt{2})}\times\dfrac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})}-2$

here the denominator part is 1,
(a+b) (a-b) = a² - b²

(3 + 2$\sqrt{2}$) \times (3 - 2$\sqrt{2}$) = 3 ² - (2$\sqrt{2})^{2}$

=9-4*2(square of $\sqrt{2}$ is 2)

=9-8

=1

= (3 + 2$\sqrt{2}$) + (3 - 2$\sqrt{2}$) - 2

= 4.

$ \therefore \left( \sqrt{x} -\dfrac{1}{\sqrt{x}} \right)$=2.

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