Question

Given x = √3 -√2 /√3 +√2, y = √3 + √2 / √3 -√2 then find x^2 + y^2 + xy y​

Answer 1

Given :-

• $\sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

• $\sf{y=\dfrac{\sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2} } }$

To Find :-

• Value of $\sf{x^2+y^2+xy}$

Solution :-

After rationalising x and y, we get

$\sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\\\\\\\sf{\implies x=\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2} )}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2} ) } }\\\\\\\sf{\implies x=\dfrac{(\sqrt{3}-\sqrt{2} )^2}{(\sqrt{3} )^2-(\sqrt{2} )^2} }\\\\\\\sf{\implies x=\dfrac{(\sqrt{3} )^2+(\sqrt{2} )^2-2\times \sqrt{3}\times \sqrt{2} }{3-2} }\\\\\\\sf{\implies x=3+2-2\sqrt{6} }\\\\\boxed{\sf{\implies x=5-2\sqrt{6} }}$

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$\sf{y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\\\\\\\sf{\implies y=\dfrac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2} )}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2} ) } }\\\\\\\sf{\implies y=\dfrac{(\sqrt{3}+\sqrt{2} )^2}{(\sqrt{3} )^2-(\sqrt{2} )^2} }\\\\\\\sf{\implies y=\dfrac{(\sqrt{3} )^2+(\sqrt{2} )^2+2\times \sqrt{3}\times \sqrt{2} }{3-2} }\\\\\\\sf{\implies y=3+2+2\sqrt{6} }\\\\\boxed{\sf{\implies y=5+2\sqrt{6} }}$

Hence,

$\sf{xy=(5-2\sqrt{6})(5+2\sqrt{6}) }\\\\\sf{\implies xy=(5)^2-(2\sqrt{6} )^2}\\\\\sf{\implies xy=25-24}\\\\\sf{\implies xy=1}$

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Now, find the value of x² and y²

$\sf{x^2=(5-2\sqrt{6})^2 }\\\\\sf{\implies x^2=(5)^2+(2\sqrt{6})^2-2\times5\times2\sqrt{6}}\\\\\sf{\implies x^2=25+24-20\sqrt{6} }\\\\\sf{\implies x^2=49-20\sqrt{6} }$

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$\sf{y^2=(5+2\sqrt{6})^2 }\\\\\sf{\implies y^2=(5)^2+(2\sqrt{6})^2+2\times5\times2\sqrt{6}}\\\\\sf{\implies x^2=25+24+20\sqrt{6} }\\\\\sf{\implies y^2=49+20\sqrt{6} }$

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Now, substitute all the values

$\sf{x^2+y^2+xy=49-20\sqrt{6}+49+20\sqrt{6}+1}\\\\\sf{\longrightarrow x^2+y^2+xy=49+49+1}\\\\\boxed{\sf{\longrightarrow x^2+y^2+xy=99}}$

Hence,

Value of x² + y² + xy is = 99

Answer 2

Step-by-step explanation:

99.................