If x = 1 + $\sqrt{2}$ $\sqrt 2$ + $\sqrt{3}$ $\sqrt 3$ and y = 1 + $\sqrt{2}$ $\sqrt 2$ - $\sqrt{3},$ $\sqrt 3 {\text{,}}$ then the value of $\frac{x^{2} + 4 x y + y^{2}}{x + y}$ $\frac{{{x^2} + 4xy + {y^2}}}{{x + y}}$ is?

1 $2 \sqrt{2}$ $2\sqrt 2$
2 $2 (2 + \sqrt{2})$ $2\left( {2 + \sqrt 2 } \right)$
31
46

Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to the question, \\ x = 1 + \sqrt{2} + \sqrt{3} . . . . . (i) \\ y = 1 + \sqrt{2} - \sqrt{3} . . . . . (i i) \\ \Rightarrow \frac{x^{2} + 4 x y + y^{2}}{x + y} \\ \Rightarrow \frac{{(x + y)}^{2} + 2 x y}{x + y} \\ From equation (i) + (ii) \\ \Rightarrow x + y = 2 + 2 \sqrt{2} \\ x y = {(1 + \sqrt{2})}^{2} - {(\sqrt{3})}^{2} \\ \Rightarrow x y = 3 + 2 \sqrt{2} - 3 \\ \Rightarrow x y = 2 \sqrt{2} \\ So, \frac{{(x + y)}^{2} + 2 x y}{x + y} \\ = \frac{{(2 + 2 \sqrt{2})}^{2} + 2 \times 2 \sqrt{2}}{2 + 2 \sqrt{2}} \\ = \frac{4 + 8 + 8 \sqrt{2} + 4 \sqrt{2}}{2 + 2 \sqrt{2}} \\ = \frac{12 + 12 \sqrt{2}}{2 + 2 \sqrt{2}} \\ = \frac{12 (1 + \sqrt{2})}{2 (1 + \sqrt{2})} \\ = 6 \end{aligned}

$\eqalign{ & {\text{According to the question,}} \cr & x = 1 + \sqrt 2 + \sqrt 3 \,.....(i) \cr & y = 1 + \sqrt 2 - \sqrt 3 \,.....(ii) \cr & \Rightarrow \frac{{{x^2} + 4xy + {y^2}}}{{x + y}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & {\text{From equation (i)}} + {\text{(ii)}} \cr & \Rightarrow x + y = 2 + 2\sqrt 2 \cr & xy = {\left( {1 + \sqrt 2 } \right)^2} - {\left( {\sqrt 3 } \right)^2} \cr & \Rightarrow xy = 3 + 2\sqrt 2 - 3 \cr & \Rightarrow xy = 2\sqrt 2 \cr & {\text{So, }}\frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & = \frac{{{{\left( {2 + 2\sqrt 2 } \right)}^2} + 2 \times 2\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{4 + 8 + 8\sqrt 2 + 4\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12 + 12\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12\left( {1 + \sqrt 2 } \right)}}{{2\left( {1 + \sqrt 2 } \right)}} \cr & = 6 \cr}$