Question

if X = √3+1/√3-1 and Y = √3-1/√3+1 then the value of x²+y² is​

Answer 1

Answer: 14

Step-by-step explanation:

We know that $x=\frac{\sqrt{3}+1 }{\sqrt{3}-1}$ and  $y=\frac{\sqrt{3}-1}{\sqrt{3}+1}$. Squaring and adding both, we get

$x^2+y^2=\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)^2}+\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)^2}$

⇒  $x^2+y^2=\frac{4+2\sqrt{3}}{4-2\sqrt{3}}+\frac{4-2\sqrt{3}}{4+2\sqrt{3}}$

Taking L. C. M., we get

$x^2+y^2=\frac{(4+2\sqrt{3})^2+(4-2\sqrt{3})^2}{(4-2\sqrt{3})(4+2\sqrt{3})}$

⇒   $x^2+y^2=\frac{(16+12+16\sqrt{3})+(16+12-16\sqrt{3})}{(16-12)}$

⇒   $x^2+y^2=\frac{56}{4}=14$

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