If $$x = 3 + 2\sqrt 2 {\text{,}}$$ then the value of $$\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right){\text{ is?}}$$
A. 1
B. 2
C. $${\text{2}}\sqrt 2 $$
D. $${\text{3}}\sqrt 3 $$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \Rightarrow x = 2 + 1 + 2\sqrt 2 \cr & \Rightarrow x = {\left( {\sqrt 2 + 1} \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 2 + 1 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 2 - 1 \cr & \therefore \sqrt x - \frac{1}{{\sqrt x }} \cr & = \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right) \cr & = \sqrt 2 + 1 - \sqrt 2 + 1 \cr & = 2 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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