Question

If $x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\frac{\sqrt{3}-1}{\sqrt{3}+1}$ find the value of $x^2+xy+y^2$

• A. 18
• B. 16
• C. 21
• D. 15

Answer 1

Answer: (D)

15

Here, we can see $x=\frac{1}{y}$

So, $x^2+xy+y^2={\left(\frac{1}{y}\right)}^2+\frac{1}{y}y+y^2=y^2+\frac{1}{y^2}+1$

Now, $y^2={\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)}^2=\frac{3+1-2\sqrt{3}}{3+1+2\sqrt{3}}=\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{(2-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=7-4\sqrt{3}$

and, $x^2=y^{-2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=7+4\sqrt{3}$

So, $x^2+xy+y^2=7-4\sqrt{3}+7+4\sqrt{3}+1=15$