If $$x = 5 + 2\sqrt 6 {\text{,}}$$ then the value of $$\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)\,{\text{is?}}$$
A. $${\text{2}}\sqrt 2 $$
B. $${\text{3}}\sqrt 2 $$
C. $${\text{2}}\sqrt 3 $$
D. $${\text{3}}\sqrt 3 $$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x = 5 + 2\sqrt 6 \cr & \Leftrightarrow x = 3 + 2 + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2} + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 + \sqrt 2 } \right)^2} \cr & \Leftrightarrow \sqrt x = \sqrt 3 + \sqrt 2 \cr & {\text{Similarly,}} \cr & \Leftrightarrow \frac{1}{{\sqrt x }} = \sqrt 3 - \sqrt 2 \cr & \therefore \left( {\sqrt x + \frac{1}{{\sqrt x }}} \right) \cr & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & = 2\sqrt 3 \cr} $$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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