If x = $\sqrt{6} + 2 and y =\sqrt{6} – 2$ then what is the value of $(\frac{ x}{y}+\frac{y}{x} )^ 2 – 3 $?
If x = $\sqrt{6} + 2 and y =\sqrt{6} – 2$ then what is the value of $(\frac{ x}{y}+\frac{y}{x} )^ 2 – 3 $?
यदि $x=\sqrt{6}+2$ तथा $y=\sqrt{6}-2$ हैं, तो $\left(\frac{x}{y}+\frac{y}{x}\right)^2-3$ का मान क्या होगा?
Detailed Solution & Logic
Correct Answer:
97
Solution
Given,
$x=\sqrt{6}+2,\quad y=\sqrt{6}-2$
First, find $xy$:
$xy=(\sqrt{6}+2)(\sqrt{6}-2)=6-4=2$
Also,
$x+y=2\sqrt{6}$
Now,
$x^2+y^2=(x+y)^2-2xy=(2\sqrt{6})^2-2(2)=24-4=20$
Therefore,
$\frac{x}{y}+\frac{y}{x}=\frac{x^2+y^2}{xy}=\frac{20}{2}=10$
Hence,
$\left(\frac{x}{y}+\frac{y}{x}\right)^2-3=10^2-3=100-3=97$
Answer: 97
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