Question

If x minus one upon x is equal to make 8 find the value of x square + 1 upon x square and X square 4 plus 1/x square 4
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Answer 1

Given

• $\sf{x - \dfrac{1}{x} = 8}$

To Find

• $\sf{ {x + \dfrac{1}{x} } }$

___________Solution___________

$~~~\implies~~~ \sf{x - \dfrac{1}{x} = 8}$

• On Squaring Both side

$~~~\implies~~~ \sf{ {\bigg(x - \dfrac{1}{x} \bigg) }^{2} = {8}^{2} }$

$~~~\implies~~~ \sf{ {x + \dfrac{1}{x} } - 2 \times x \times \dfrac{1}{x} = 64 }$

$~~~\implies~~~ \sf{ {x + \dfrac{1}{x} } = 64 + 2 }$

$~~~\implies~~~ \sf{ {x + \dfrac{1}{x} } = 66 }$

Answer 2

$\huge\fcolorbox{black}{magenta}{question☺᭄✍︎}$

$\tt\red{x - \dfrac{1}{x} = 8}$

$\huge\pink{\boxed{\green {\mathbb{\overbrace {\underbrace{\fcolorbox{red}{aqua}{\underline{\pink{†◖ \: ⚆to \: find⚆◗†}}}}}}}}}$

$\tt \pink{ {x + \dfrac{1}{x} } }$

$\huge\fcolorbox{pink}{Red}{ ★A᭄ΠSWΣR✍︎}$

$~~\implies~~~ \tt \orange{x - \dfrac{1}{x} = 8}$

On Squaring Both side

$~~\implies~~~ \tt\purple{ {\bigg(x - \dfrac{1}{x} \bigg) }^{2}}$

$~~\implies~~~ \tt \green{ {x + \dfrac{1}{x} } - 2 \times x \times \dfrac{1}{x} = 64 }$

$~~\implies~~~ \tt\pink{ {x + \dfrac{1}{x} } = 64 + 2 }$

$~~\implies~~~ \tt\red{ {x + \dfrac{1}{x} } = 66 }$

$\huge\pink{\mid{\fbox{\tt{thanks}}\mid}}$