If x = 3 + 2$$\sqrt 2 $$ and xy = 1, then the value of $$\frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}}$$ is?
A. $$\frac{{30}}{{31}}$$
B. $$\frac{{70}}{{31}}$$
C. $$\frac{{35}}{{31}}$$
D. $$\frac{{37}}{{31}}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & x = 3 + 2\sqrt 2 {\text{ and }}xy = 1 \cr & {y^2} = \frac{1}{{{x^2}}} \cr & y = \frac{1}{x} = \frac{1}{{3 + 2\sqrt 2 }} = 3 - 2\sqrt 2 \cr & \therefore x + \frac{1}{x} = 3 + 2\sqrt 2 + 3 - 2\sqrt 2 = 6 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 36 - 2 = 34 \cr & \frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}} \cr & = \frac{{{x^2} + \frac{1}{{{x^2}}} + 3}}{{{x^2} + \frac{1}{{{x^2}}} - 3}} \cr & = \frac{{34 + 3}}{{34 - 3}} \cr & = \frac{{37}}{{31}} \cr} $$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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