Question

if a is equals to root 3 + root 2 upon root 3 minus root 2 and b is equals to root 3 minus root 2 upon root 3 + root 2 find the value of a square + b square​

Answer 1

Answer:

a =  \frac{ \sqrt{3} -  \sqrt{2}  }{ \sqrt{3} +  \sqrt{2}  }  \\  \\ b =  \frac{ \sqrt{3} +  \sqrt{2}  }{ \sqrt{3}  -  \sqrt{2} }  \\  \\ ab =  \frac{( \sqrt{3}  -  \sqrt{2} )}{ (\sqrt{ 3}  +  \sqrt{2} )}  \frac{( \sqrt{3}  +  \sqrt{2}) }{( \sqrt{3} -  \sqrt{2} ) }  \\  \\ ab = 1 \\  \\ now \\  \\ a =  \frac{( \sqrt{3} -  \sqrt{2})  }{( \sqrt{3}  +  \sqrt{2} )}  \\  \\ a =  \frac{ (\sqrt{3}  -  \sqrt{2} ) }{ (\sqrt{3} +  \sqrt{2})  }  \frac{ (\sqrt{3} -  \sqrt{2} ) }{( \sqrt{3}   -   \sqrt{2} )}  \\  \\ a =  {( \sqrt{3} -  \sqrt{2} ) }^{2}  \\  \\ a = 5 - 2 \sqrt{6}  \\  \\ b =  \frac{( \sqrt{3} +  \sqrt{2}  )}{( \sqrt{3}  -  \sqrt{2} )}  \\  \\ b =  \frac{( \sqrt{3}  +  \sqrt{2} )}{( \sqrt{3} -  \sqrt{2})  }   \frac{( \sqrt{3} +  \sqrt{2}  )}{( \sqrt{3}  +  \sqrt{2}) }  \\  \\ b =  {( \sqrt{3} +  \sqrt{2})  }^{2}  \\  \\ b = 5 + 2 \sqrt{6}  \\  \\  {a}^{2}  = 49  - 20 \sqrt{6}  \\  \\  {b}^{2}  = 49 + 20 \sqrt{6}  \\  \\ now \\  \\  {a}^{2}  +  {b}^{2}  - 5ab \\  \\  = 93

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Explanation: