Question

If x=√3+√2/√3-√2 and y=√3-√2/√3+√2 find the value of x^2+y^2

Answer 1

Answer:

x²+y²=98

Step-by-step explanation:

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Answer 2

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ xy = 1 \\ \\ x + y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x + y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} + ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x + y = 2(3 + 2) = 10 \\ \\ \\ {x}^{2} + {y}^{2} = ( {x + y})^{2} - 2xy \\ = {10}^{2} - 2\\ = 98$