Question

14. If x + 1 = 13, then the value of x...
pls answer fast!​

Answer 1

GIVEN :-

• x + 1/x = 13.

TO FIND :-

• The value of (x - 1/x).

SOLUTION :-

Firstly we will find the value of x² + 1/x².

$\implies \displaystyle \sf \:x + \frac{1}{x} = 13$

On squaring both the sides we get,

$\implies \displaystyle \sf \: \bigg \lgroup x + \frac{1}{x} \bigg \rgroup ^{2} = (13) ^{2}$

$\implies \displaystyle \sf \:x ^{2} + \frac{1}{x ^{2} } + 2 \times x ^{2} \times \frac{1}{x ^{2} } = 169$

$\implies \displaystyle \sf \:x ^{2} + \frac{1}{x ^{2} } + 2 = 169$

$\implies \displaystyle \sf \:x ^{2} + \frac{1}{x ^{2} } = 169 - 2$

$\implies \displaystyle \sf \:x ^{2} + \frac{1}{x ^{2} } = 167$

Now according to the given question,

$\implies \displaystyle \sf \: \bigg \lgroup x - \frac{1}{x } \bigg \rgroup ^{2} = x ^{2} + \frac{1}{x ^{2} } - 2 \times x ^{2} \times \frac{1}{x ^{2} }$

$\implies \displaystyle \sf \: \bigg \lgroup x - \frac{1}{x } \bigg \rgroup ^{2} =167 - 2$

$\implies \displaystyle \sf \: \bigg \lgroup x - \frac{1}{x } \bigg \rgroup ^{2} =165$

$\implies \underline{ \boxed{\displaystyle \sf \:x - \frac{1}{x} = \pm \sqrt{165} }}$

Hence the correct option is (E) none of the above.

Answer 2

Answer:

$\sf\huge\red{Given}$

$\sf{x \: + \: \frac{1}{x} = 13}$

$\sf\huge\red{To \:find}$

$\sf\large\red{The \: value \: of - \: x - \frac{1}{x } }$

$\sf{Firstly \: we \: will \: find \: the \: value \: of \: x²+\frac{1}{ {x}^{2} } }$

$\sf{x \: + \: \frac{1}{x} = 13}$

$\sf{on \: squaring \: both \: side =>}$

$\sf{( {x}+ \frac{1}{ {x} }) ^{2} = {(13)}^{2} } \sf$

$\sf{ ( {x}+ \frac{1}{ {x} }) ^{2} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } = 169 }$

$\sf{( {x}^{2} + \frac{1}{ {x ^{2} } }) = 169 - 2}$

$\sf{ {x}^{2} + \frac{1}{ {x}^{2} } = 167}$

$\sf\red{According \: to \: Question}$

$\sf{ {(x - \frac{1}{x}) }^{2} = {x}^{2} + {( \frac{1}{x}) }^{2} } - 2 \times {x}^{2} \times \frac{1}{ {x}^{2} }$

$\sf{ {(x - \frac{1}{x}) }^{2} = 167 - 2} \\ \sf{{(x - \frac{1}{x}) }^{2} = 165} \\ \sf{x - \frac{1}{x} = \sqrt{165} }$

$\sf{So, option \: C \: is \: correct}$