Math, asked by Anonymous, 2 months ago

If x = 1
√2+√3
and y = 1
√2− √3
, find the value of x2 + xy + y2

Answers

Answered by TYKE
5

Question :

 \sf \small \maltese  \: if  \: x =  \frac{1}{ \sqrt{2} +  \sqrt{3}  }  \: and \: y =  \frac{1}{ \sqrt{2} -  \sqrt{3}  }  \: then \: find \: the \: value \: of \:  {x}^{2}  + xy +  {y}^{2}

Solution :

First we need to rationalise the denominator in both x and y then simplify it according to the given condition

 \sf \small \dashrightarrow x =  \frac{1}{ \sqrt{2}  +  \sqrt{3} }

  \sf \small \dashrightarrow x =  \frac{1( \sqrt{2} -  \sqrt{3})  }{( \sqrt{2} +  \sqrt{3} )( \sqrt{2}  -  \sqrt{3})  }

 \sf \small \dashrightarrow x =  \frac{ \sqrt{2}  -  \sqrt{3} }{ {( \sqrt{2} )}^{2} -  {( \sqrt{3}) }^{2}  }

 \sf \small \dashrightarrow x =  \frac{ \sqrt{2} -  \sqrt{3}  }{2 - 3}

 \sf \small \dashrightarrow x =   -  \sqrt{2}  +  \sqrt{3}

Now to rationalize y

 \sf \small \dashrightarrow y =  \frac{1}{ \sqrt{2}   -  \sqrt{3} }

  \sf \small \dashrightarrow y =  \frac{1( \sqrt{2}  +  \sqrt{3})  }{( \sqrt{2}  -   \sqrt{3} )( \sqrt{2}   +  \sqrt{3})  }

 \sf \small \dashrightarrow y=  \frac{ \sqrt{2}   +  \sqrt{3} }{ {( \sqrt{2} )}^{2} -  {( \sqrt{3}) }^{2}  }

 \sf \small \dashrightarrow y=  \frac{ \sqrt{2}  +  \sqrt{3}  }{2 - 3}

 \sf \small \dashrightarrow y =   -  \sqrt{2}   -   \sqrt{3}

Finally we rationalized both x and y

Now to simplify the given condition

 \sf \small \dashrightarrow  {x}^{2}  + xy +  {y}^{2}

Putting the values we get

 \sf \small \dashrightarrow ( -  \sqrt{2}  +   \sqrt{3} )^{2}  + ( -  \sqrt{2}   +  \sqrt{3} )( -  \sqrt{2}  -  \sqrt{3} ) +  {( -  \sqrt{2}  -  \sqrt{3} )}^{2}

 \sf \tiny \dashrightarrow  {( -  \sqrt{2} )}^{2}  + 2 ( -  \sqrt{2} )( \sqrt{3} ) +  {( \sqrt{3} )}^{2}   +  {(  -\sqrt{2} )}^{2} - {( \sqrt{3}) }^{2}  + ( -  \sqrt{2} )^{2}  - 2( -  \sqrt{2} )( -  \sqrt{3} ) +  {(  - \sqrt{3} )}^{2}

 \sf \small \dashrightarrow 2 - 2 \sqrt{6}   + 3 + 2 - 3 + 2 + 2 \sqrt{6}  + 3

 \sf \small \dashrightarrow 2 + 3 -  \cancel{2 \sqrt{6}} + 2 + 3 +  \cancel{2 \sqrt{6} } - 1

 \sf \small \dashrightarrow 5 + 5 - 1

 \sf \small \dashrightarrow 10 - 1

 \sf \small \dashrightarrow 9

So the value of x² + xy + y² is 9

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