Question : If $x=\frac{1}{2+\sqrt{3}}$ and $y=\frac{1}{2-\sqrt{3}},$ then the value of $8xy(x^{2}+y^{2})$ is:
Option 1: 196
Option 2: 290
Option 3: 112
Option 4: 194
Correct Answer: 112
Solution : To solve this problem, we first need to calculate the values of $x$ and $y$. $x=\frac{1}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{1}$ $y=\frac{1}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{1}$ Now, $x+y =4$............(I) Also, $xy=(2-\sqrt{3})\times (2+\sqrt{3}) =1$.............(II) Squaring both sides of equation (I) ⇒ $x^2+y^2+2xy=16$ ⇒ $x^2+y^2+2\times1=16$ ⇒ $x^2+y^2=14$ Putting the value of $x^2+y^2$ and $xy$ in $8xy(x^{2}+y^{2})$, we have, ⇒ $8xy(x^{2}+y^{2})= 8 \times 1 (14) =112$ Hence, the correct answer is 112.
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