X - - 3 - 2 and y - 2 - - 3 then find value of x 2 - y 2

Given x=√(3) 2,y =2+√(3) x^2+y^2=(√(3) 2)^2+(2+√(3))^2 #160; #160; #160; #160; #160; #160; #160; =((√(3))^2+(2)^2 2(√(3))(2) ...

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Addition of Polynomials

x = √ 3 - 2 a...

Question

$x = \sqrt{3} - 2$ and $y = 2 + \sqrt{3}$ then find value of $x^{2} + y^{2}$

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x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}

y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}

x^{2} + x y - y^{2}

x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}; y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}

x^{2} + y^{2}

x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}

y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}

x^{2} + y^{2}

x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} a n d y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}

x^{2} + y^{2}

\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}},

\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}},

x^{2} + x y - y^{2} .

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