Question

if x=√3+1/√3-1,Y=√3-1/√3+1,than find the value of x^2+xy+y^2

Answer 1

Identities used :

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

We are given the values :

$x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 }$

$y = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 }$

Since both the denominator are not rational, let us first rationalize it :

$x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } \times \frac{ \sqrt{3} + 1 }{ \sqrt{3} + 1} \\ \\ x = \frac{ {( \sqrt{3} + 1 )}^{2} }{( \sqrt{3} - 1)( \sqrt{3} + 1) } \\ \\ x = \frac{ {( \sqrt{3} })^{2} + {(1)}^{2} + 2( \sqrt{3} )(1) }{ {( \sqrt{3} })^{2} - {(1)}^{2} } \\ \\ x = \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} \\ \\ x = \frac{4 + 2 \sqrt{3} }{2} \\ \\ x = 2 + \sqrt{3}$

Similarly,

$y = 2 - \sqrt{3}$

Now, Putting the values in the given equation :

${x}^{2} + xy + {y}^{2} \\ \\ = {(2 + \sqrt{3}) }^{2} + (2 + \sqrt{3} )(2 - \sqrt{3} ) + {(2 - \sqrt{3} )}^{2} \\ \\ = ( {(2)}^{2} + {( \sqrt{3} )}^{2} + 2(2)( \sqrt{3} )) \\ + ( {(2)}^{2} - {( \sqrt{3}) }^{2} ) + ( {(2)}^{2} + {( \sqrt{3}) }^{2} - 2(2)( \sqrt{3}) ) \\ \\ = (4 + 3 + 4 \sqrt{3} ) + (4 - 3) + (4 + 3 - 4 \sqrt{3} ) \\ \\ = 7 + 4 \sqrt{3} + 1 + 7 - 4 \sqrt{3} \\ \\ = 7 + 1 + 7 \\ \\ \bf = 15$