If $$x = 7 - 4\sqrt 3 {\text{,}}$$ then $$\sqrt x {\text{ + }}\frac{1}{{\sqrt x }}$$ is equal to?
A. 1
B. 2
C. 3
D. 4
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & x = 7 - 4\sqrt 3 \cr & \Rightarrow x = 4 + 3 - 4\sqrt 3 \cr & \Rightarrow x = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2 \times 2\sqrt 3 \cr & \Rightarrow {\left( {2 - \sqrt 3 } \right)^2} \cr & \therefore \left[ {{a^2} + {b^2} - 2ab = {{\left( {a - b} \right)}^2}} \right] \cr & \Rightarrow x = {\left( {2 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + \sqrt 3 \cr & \therefore \sqrt x {\text{ + }}\frac{1}{{\sqrt x }}{\text{ }} \cr & = 2 - \sqrt 3 + 2 + \sqrt 3 \cr & = 4 \cr} $$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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