Discussion :: Square Root and Cube Root
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If x =\(\frac { \sqrt{3} +1} {\sqrt{3}-1 } \) and Y = \(\frac { \sqrt{3} -1} {\sqrt{3}+1 } \) ,then the value of (x2 + y2) is:
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Answer : Option C
Explanation :
X = \(\frac {( \sqrt{3} +1)} {(\sqrt{3}-1) } \) x \(\frac { (\sqrt{3} +1)} {(\sqrt{3}+1 )} \) = \(\frac { (\sqrt{3} +1)^2} {({3}-1 )} \) = \(\frac { 3+1+2\sqrt{3} } {2 } \) = 2+\( \sqrt{3}\)
y = \(\frac {( \sqrt{3} -1)} {(\sqrt{3}+1) } \) x \(\frac { (\sqrt{3} -1)} {(\sqrt{3}-1 )} \) = \(\frac { (\sqrt{3}-1)^2} {({3}-1 )} \)= \(\frac { 3+1-2\sqrt{3} } {2 } \)= 2 -\( \sqrt{3}\)
x2 + y2 = (2 +\( \sqrt{3}\) )2 + (2 - \( \sqrt{3}\))2
= 2(4 + 3)
= 14
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