Question

If x^²+1/2x^²=27 find x + 1/x​

Answer 1

your answer mate

just expand, rearrange n substitute,!

(x−1x)2

=x2−2⋅x⋅1x+(1x)2

=x2−2+(1x)2

=x2+1x2−2

since we have the value of x2+1x2 as 27,

=27−2

=25

⇒(x−1x)2=25

⇒x−1x=±5

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Answer 2

$\bold{Correct\: Question:}$

${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 27$

Solution:

${x}^{2} \: + \: \dfrac{1}{{x}^{2}} \: = \: 27$

We know that..

a² + b² = (a + b)² - 2ab

⟹ ${ \bigg(x \: + \: \dfrac{1}{x} \bigg)}^{2}$ - $2{ \cancel{x}} \: \times \: \dfrac{1}{ \cancel{x}}\:=\:27$

⟹ ${ \bigg(x \: + \: \dfrac{1}{x} \bigg)}^{2}$ - $2\:=\:27$

⟹ ${ \bigg(x \: + \: \dfrac{1}{x} \bigg)}^{2}$ = $27\:+\:2$

⟹ ${ \bigg(x \: + \: \dfrac{1}{x} \bigg)}^{2}$ = 29

⟹ $x \: + \: \dfrac{1}{x}$ = $\sqrt{29}$

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$x \: + \: \dfrac{1}{x}$ = $\sqrt{29}$

___________ [ANSWER]

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