Math, asked by kallapadmakumari, 8 months ago

x²-5x+1=0 then x²+1/x²=​

Answers

Answered by Anonymous
2

Answer:

x²-5x+1=0

x²+1=5x

dividing both side by x

x²/x+ 1/ x= 5x/x

x+1/x=5

your answer is 5

Answered by Anonymous
4

\large  \sf\underline{ \underline{ \red{given : }}} \\  \\

 \tt{ {x}^{2}  - 5x + 1 = 0}

 \\  \\ \large  \sf\underline{ \underline{ \red{to \: find : }}} \\  \\  \tt{ {x}^{2} +  \frac{1}{ {x}^{2} }  }

 \\  \\ \large  \sf\underline{ \underline{ \red{solution : }}} \\  \\

We know ,

 \\  \sf{ {x}^{2} - 5x + 1 = 0 } \\  \\  \\  \rm{dividing \: whole \: equation \: by \: x} \\   \\ \\  \sf{  \frac{ { {x}^{ \cancel2  \:  1} } }{ \cancel{x}}  -  \frac{5 \cancel{x}}{ \cancel{x}}  +  \frac{1}{x}  =  \frac{0}{x} } \\  \\  \\  \sf{x - 5 +  \frac{1}{x} = 0 } \\  \\  \\  \sf{x +  \frac{1}{x}  = 5} \\ \\   \\  \rm{squaring \: both \: sides...} \\  \\  \sf{(x +  { \frac{1}{x} })^{2} } =  {5}^{2}  \\   \\  \boxed{ \sf{( {a + b)}^{2}  =  {a + 2ab +  {b}^{2} }}} \\

Here ,

  • a = x

  • b = 1/x

Putting values , we get..

 \\  \sf{ {x}^{2} + 2( \cancel{x})( \frac{1}{ \cancel{x} } ) + ( { \frac{1}{x} )}^{2} } =  {5}^{2}  \\  \\  \\ \sf{ {x}^{2} + 2 +  \frac{1}{ {x}^{2} }   = 25} \\  \\  \\  \boxed   {\sf{  \blue{{x}^{2} +  \frac{1}{ {x}^{2}  }  = 23 }}}

 \\  \\

OTHER IMPORTANT IDENTITIES :-

  • ( a - b )² = a² - 2ab + b²

  • ( a + b ) ( a - b ) = a² - b²

  • ( a + x ) ( a + y ) = a² + ( x + y )a + xy
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