Question

if
${x}^{2} + 1 \div x = 66 \: find \: the \: value \: of \\ x - 1 \div x$
​

Answer 1

Given that :

$\frac{ {x}^{2} + 1 }{x} = x + \frac{1}{x} = 66$

As we know that :

${(a - b)}^{2} = {(a + b)}^{2} - 4ab$

Such as :

${(x - \frac{1}{x} )}^{2} = {(x + \frac{1}{x} )}^{2} - 4 \times x \times \frac{1}{x} \\ \\ = > {(x - \frac{1}{x} )}^{2} = {(66)}^{2} - 4 \\ \\ = > {(x - \frac{1}{x}) }^{2} = 4356 - 4 = 4352 \\ \\ = > x - \frac{1}{x} = 65.9 = 66 \: (approx)$

@ItsChampion ✌️

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

$\sf{ {x}^{2} + \frac{1}{ {x}^{2} } = 66}$

$\bf \underline{To find-}$

$\sf{the \: value \: of :\: x - \frac{1}{x} = \: ?}$

$\bf \underline{Solution-}$

$\sf {\bigg(x - \frac{1}{x} \bigg) ^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \: \: \: \: \: \: [ \because \: (a - b {)}^{2} = {a}^{2} + {b}^{2} - 2ab ]} \\ \\ = 66 - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\rm{ [ \because {x}^{2} + \frac{1}{ {x}^{2}} = 66 \:(Given) ] }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= 64$

$\sf{ \therefore \: \: \: \: \: \: x - \frac{1}{x} = \sqrt{64} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = ± \: 8$

$\bf\underline{Hence,the \: value \: of : \: x - \frac{1}{x} \: is \: ± \: 8.}$