Question

If x= √3+1/√3-1
And y= √3-1/√3+1
Then find x²+y²?

Answer 1

Answer:

2

Step-by-step explanation:

We know that,

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

So,

${x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy$

So,

${x}^{2} + {y}^{2}$

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

So,

${x}^{2} + {y}^{2}$

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

$= \frac{8}{2} - 2 = 4 - 2 = 2$

So,

${x}^{2} + {y}^{2} = 2$