If x-3-23 -2and y-3-23-2- find x2-y2-

X #160;=3+23 2 #215;3+23+2=3+223 2=3+2+223=5+26Now,y #160;=3 23+2 #215;3 23 2=3 223 2=3+2 223=5 26Thus, #160;we #160;have:xy=5+265 26=25 24=1x+y=5+2 ...

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Byju's Answer

Standard VIII

Mathematics

Laws of Exponents

If x=3+23 -2a...

Question

If $x = \frac{(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})} and y = \frac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})}$ , find (x² + y²).

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Solution

$x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{{(\sqrt{3} + \sqrt{2})}^{2}}{3 - 2} = 3 + 2 + 2 \sqrt{2} \sqrt{3} = 5 + 2 \sqrt{6} Now, y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{{(\sqrt{3} - \sqrt{2})}^{2}}{3 - 2} = 3 + 2 - 2 \sqrt{2} \sqrt{3} = 5 - 2 \sqrt{6} Thus, we have : x y = (5 + 2 \sqrt{6}) (5 - 2 \sqrt{6}) = 25 - 24 = 1 x + y = (5 + 2 \sqrt{6}) + (5 - 2 \sqrt{6}) \Rightarrow x + y = 10 Squaring both sides, we get : {(x + y)}^{2} = 10^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 100 \Rightarrow x^{2} + y^{2} + 2 \times 1 = 100 \Rightarrow x^{2} + y^{2} = 100 - 2 = 98$

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If x=3+23-2and y=3-23+2, find (x2 + y2).

x =3+23-2×3+23+2=3+223-2=3+2+223=5+26Now,y =3-23+2×3-23-2=3-223-2=3+2-223=5-26Thus, we have:xy=5+265-26=25-24=1x+y=5+26+5-26⇒x+y=10Squaring both sides, we get:x+y2=102⇒x2+y2+2xy=100⇒x2+y2+2×1=100⇒x2+y2=100-2=98

If $x = \frac{(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})} and y = \frac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})}$ , find (x² + y²).