Question

x²-5x+1=0 then x²+1/x²=​

Answer 1

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

Answer 2

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

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

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

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

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