Math, asked by jhanavi61270, 11 months ago

If x=√3+√2/√3-√2 find x^2+1/x^2

Answers

Answered by Anonymous
3

given us ....

x=(3+2)/3-2.

=(3+2)^2.

=3+2+26.

=5+26.

so..

(x^2+1)/x^2=

put value of x...

{(5+26)^2+1}/(5+26)

=(25+24+206+1)/5+26)

=(50+206)/(5+26)

=10.

i hope this is right and helpful for u.

Answered by Anonymous
9

Given: \:\sf x =  \frac{ \sqrt{3} +  \sqrt{2}  }{ \sqrt{3}  -  \sqrt{2} }

Rationalise the denominator

 \sf x =  \frac{ \sqrt{3} +  \sqrt{2}  }{ \sqrt{3}  -  \sqrt{2} }  \times  \frac{ \sqrt{3}  +  \sqrt{2} }{ \sqrt{3}  +  \sqrt{2} }

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

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

 \sf x =  \frac{ 3  +  {2}  + 2\sqrt{6}  }{3  -  2 }

 \sf x =  { 5  + 2\sqrt{6}  }

Now,

  \sf\frac{1}{x}  =  \frac{1}{5 + 2 \sqrt{6} }

rationalise the denominator.

  \sf\frac{1}{x}  =  \frac{1}{5 + 2 \sqrt{6} }  \times  \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} }

 \sf\frac{1}{x}  =  \frac{5 - 2 \sqrt{6} }{ {5}^{2}  - {( 2 \sqrt{6} ) }^{2} }

 \sf\frac{1}{x}  =  \frac{5 - 2 \sqrt{6} }{ {25 - 24 }}

 \sf\frac{1}{x}  =  5 - 2 \sqrt{6}

 \sf {x}^{2}  +   \frac{1}{ {x}^{2} }  =  {(5 + 2 \sqrt{6}) }^{2}  +  {(5 - 2 \sqrt{6} )}^{2}

 \fbox{ \sf {x}^{2}  +   \frac{1}{ {x}^{2} }  =  98}

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