Question

if x=√3-√2/√3+√2 and y = √3+√2 /√3-√2 , find the value of x2 + y2 + x y​

Answer 1

Step-by-step explanation:

let x=√3 -√2/√3 + √2

let y=√3 +√2/√3-√2

then x²=(3+2-2√6)/3+2+2√6

x²=(5-2√6)/(5+2√6)

y²=(5+2√6)/(5-2√6)

xy=1

x²+y²+xy=(5-2√6/5+2√6)+(5+2√6/5-2√6)+1

=>(5-2√6)²+(5+2√6)²+(5+2√6)(5-2√6)

=>(25+4(6)-20√6+25+4(6)+20√6)+25-24

=>>(25+24+25+24)+(25-24)

=>>(98+1)

=>x²+y²+xy =99

Answer 2

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ xy = 1 \\ \\ x + y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x + y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} + ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x + y = 2(3 + 2) = 10 \\ \\ \\ {x}^{2} + {y}^{2} + xy = ( {x + y})^{2} - xy \\ = {10}^{2} - 1 \\ = 99$