If x = 1 + $$\sqrt 2 $$ + $$\sqrt 3 $$ and y = 1 + $$\sqrt 2 $$ - $$\sqrt 3 {\text{,}}$$ then the value of $$\frac{{{x^2} + 4xy + {y^2}}}{{x + y}}$$ is?
A. $$2\sqrt 2 $$
B. $$2\left( {2 + \sqrt 2 } \right)$$
C. 1
D. 6
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{According to the question,}} \cr & x = 1 + \sqrt 2 + \sqrt 3 \,.....(i) \cr & y = 1 + \sqrt 2 - \sqrt 3 \,.....(ii) \cr & \Rightarrow \frac{{{x^2} + 4xy + {y^2}}}{{x + y}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & {\text{From equation (i)}} + {\text{(ii)}} \cr & \Rightarrow x + y = 2 + 2\sqrt 2 \cr & xy = {\left( {1 + \sqrt 2 } \right)^2} - {\left( {\sqrt 3 } \right)^2} \cr & \Rightarrow xy = 3 + 2\sqrt 2 - 3 \cr & \Rightarrow xy = 2\sqrt 2 \cr & {\text{So, }}\frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & = \frac{{{{\left( {2 + 2\sqrt 2 } \right)}^2} + 2 \times 2\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{4 + 8 + 8\sqrt 2 + 4\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12 + 12\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12\left( {1 + \sqrt 2 } \right)}}{{2\left( {1 + \sqrt 2 } \right)}} \cr & = 6 \cr} $$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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