If $$x - \frac{1}{x} = 4{\text{,}}$$ then $$\left( {x + \frac{1}{x}} \right)$$ is equal to?
A. $${\text{5}}\sqrt 2 $$
B. $${\text{2}}\sqrt 5 $$
C. $${\text{4}}\sqrt 2 $$
D. $${\text{4}}\sqrt 5 $$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x - \frac{1}{x} = 4 \cr & \left( {{\text{on squaring }}} \right) \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 - 2 = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 20 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 20 \cr & \Rightarrow x + \frac{1}{x} = \sqrt {20} \cr & \Rightarrow x + \frac{1}{x} = \sqrt {4 \times 5} \cr & \Rightarrow x + \frac{1}{x} = 2\sqrt 5 \cr} $$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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