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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