Question

x-1/x =8then what will be x²+1/x²​

Answer 1

Given:

• x - 1/x = 8

To find:

• x² + 1/x²

Solution:

Consider the given equation

$\rm{x + \dfrac{1}{x} = 8}$

Squaring on both sides,

$\rm{ \bigg(x + \dfrac{1}{x} \bigg)^{2} = 8 ^{2} }$

Split using (a + b)² = a² + b² + 2ab

$\implies \sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } + \bigg(2( \not{x}) \dfrac{1}{ \not{x}} \bigg) = 64}$

Now we have,

$\implies \rm{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 =64 }$

Transpose 2 to R.H.S

$\hookrightarrow \: \rm{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 64 - 2 }$

$\therefore \boxed{ \bf{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 62 }}$

Know more:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$