Question

alt="\huge\sf\underline{Question}" align="absmiddle" class="latex-formula">
$\sf If \: {x}^{2} + \frac{1}{x ^{2} } = 34$
$\sf find \: the \: value \: of \: \: \: x + \frac{1}{x}$
a proper explanation needed :)

Thxx !! ​

Answer 1

${\qquad\quad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

$\qquad \sf \dashrightarrow \: {x}^{2} + \cfrac{1}{ {x}^{2} } = 34$

[ add 2 on both sides ]

$\qquad \sf \dashrightarrow \: {x}^{2} + \cfrac{1}{ {x}^{2} } + 2 = 34 + 2$

[ form identity : a² + b² + 2ab ]

$\qquad \sf \dashrightarrow \: {(x)}^{2} + { \bigg(\cfrac{1}{ {x}^{} } \bigg) }^{2} + 2 \sdot(x) \sdot \bigg(\cfrac{1}{x} \bigg) = 36$

[ a² + b² + 2ab = (a + b)² ]

$\qquad \sf \dashrightarrow \: {\bigg (x + \cfrac{1}{x} \bigg) }^{2} = 36$

$\qquad \sf \dashrightarrow \: {\bigg (x + \cfrac{1}{x} \bigg) }^{} = \sqrt{ 36}$

$\qquad \sf \dashrightarrow \: x + \cfrac{1}{x} = \pm 6$

so, the value of required expression is 6

[usually positive value is considered, but if asked the value can be either positive or negative]