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  • if x = sqroot3+sqroot2 by sqroot 3-sqroot2 and y = sqroot3-sqroot2 by sqroot3+sqroot2 then the value of x pwer 2 + y power 2

If  x= \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} and y= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} then find the value of x2 + y2.

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Answer.  98
Solution.We use the identity \left ( a+b \right )^{2}= a^{2}+b^{2}+2ab
So, \left ( \sqrt{a}+\sqrt{b} \right )^{2}= a+2\sqrt{ab}+b
x^{2}+y^{2}= \left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )^{2}+\left ( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \right )^{2}
= \frac{3+2\sqrt{6}+2}{3-2\sqrt{6}+2}+\frac{3-2\sqrt{6}+2}{3+2\sqrt{6}+2}
= \frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\frac{5-2\sqrt{6}}{5+2\sqrt{6}}
= \frac{\left ( 5+2\sqrt{6} \right )^{2}+\left ( 5-2\sqrt{6} \right )^{2}}{\left ( 5-2\sqrt{6} \right ){\left ( 5+2\sqrt{6} \right )}}
 = \frac{\left ( 25+20\sqrt{6}+24 \right )+\left ( 25-20\sqrt{6} +24\right )}{25-24}

Using   (a – b) (a + b) = a2– b2= 98
Hence the answer is 98.

 

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infoexpert27

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