Question

If x = (sqrt(3) - sqrt(2))/(sqrt(3) + sqrt(2)) and y= sqrt 3 + sqrt 2 sqrt 3 - sqrt 2 , then x ^ 2 + xy + y ^ 2 =

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

If $x = \frac{\sqrt{3} -\sqrt{2} }{\sqrt{3} + \sqrt{2} }$  and  $y = \frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} }$  then the value of x² + xy + y² = 99

To find:

Given:

$x = \frac{\sqrt{3} -\sqrt{2} }{\sqrt{3} + \sqrt{2} }$  and  $y = \frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} }$  

To find:

Find the value of x² + xy + y²

Solution:

From Given data $x = \frac{\sqrt{3} -\sqrt{2} }{\sqrt{3} + \sqrt{2} }$  

To rationalize x, multiply and divide x with (√3-√2)

$x = \frac{\sqrt{3} -\sqrt{2} }{\sqrt{3} + \sqrt{2} }\frac{(\sqrt{3} - \sqrt{2} )}{(\sqrt{3}- \sqrt{2}) }$  

$x= \frac{(\sqrt{3} -\sqrt{2} )^{2} }{(\sqrt{3})^{2} - (\sqrt{2})^{2} }$

$x= \frac{(\sqrt{3})^{2} +(\sqrt{2} )^{2} - 2(\sqrt{3})(\sqrt{2}) }{(3 - 2) }$    [  Use (a - b)² = a²+b² - 2ab ]

$x =(3 +2 - 2(\sqrt{6}) }$  

$x =5- 2\sqrt{6} }$

Given  $y = \frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} }$  

To rationalize Y, multiply and divide x with (√3+√2)  

$y = \frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} }\frac{(\sqrt{3} +\sqrt{2})}{(\sqrt{3} +\sqrt{2})}$

$y = \frac{(\sqrt{3} +\sqrt{2})^{2} }{(\sqrt{3})^{2} -(\sqrt{2})^{2} }$

$y = \frac{(\sqrt{3})^{2} +(\sqrt{2})^{2} +2(\sqrt{3} )(\sqrt{2}) }{(3 -2) }$

$y = {3 +2+2\sqrt{6} }$

$y = {5+2\sqrt{6} }$

Now calculate x² + xy + y² using above values

⇒ $(5- 2\sqrt{6} )^{2}$ +$(5- 2\sqrt{6} )( {5+2\sqrt{6} })$ + $( {5+2\sqrt{6}})^{2}$

⇒  $25+24 - 20\sqrt{6} + 25- 24 + 25+24 +20\sqrt{6}}$

⇒ $25+24 + 25+ 25$

⇒ 99

Therefore, the value of x² + xy + y² = 99

 

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