Question

if x+(1/x) =1 then find the value of x^29+(1/x^89) ​

Answer 1

Answer:

x2+x−1=0⇒ x = 0 does not satisfy the equation.

⇒x2+x−1x=0, x ≠ 0

⇒x+1−1x=0

⇒x−1x=−1

Squaring,x2+1x2−2⋅x⋅1x=1

⇒x2+1x2=1+2=3

Squaring,x4+1x4+2⋅x2⋅1x2=32=9

⇒x4+1x4=9−2=7

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x + \dfrac{1}{x} = 1$

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

$\rm :\longmapsto\: {x}^{2} + 1 = x$

$\rm :\longmapsto\: {x}^{2} - x + 1 = 0$

Its a quadratic in x, so using Quadratic Formula, we have

$\rm :\longmapsto\:x = \dfrac{ - ( - 1) \: \pm \: \sqrt{ {( - 1)}^{2} - 4(1)(1)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ 1\: \pm \: \sqrt{ {1 - 4}} }{2}$

$\rm :\longmapsto\:x = \dfrac{ 1\: \pm \: \sqrt{ { - 3}} }{2}$

$\rm :\longmapsto\:x = \dfrac{ 1\: \pm \: \sqrt{ {3}} \: i }{2}$

So,

can have values,

$\rm :\longmapsto\:x = \dfrac{ 1\: + \: \sqrt{ {3}} \: i }{2} \: or \: \dfrac{ 1\: - \: \sqrt{ {3}} \: i }{2}$

We know that,

$\rm :\longmapsto\: \omega \: = \dfrac{ - 1\: + \: \sqrt{ {3}} \: i }{2} \: \\ \\ \rm :\longmapsto\: {\omega}^{2} = \: \dfrac{ - 1\: - \: \sqrt{ {3}} \: i }{2}$

So,

values of x can be rewritten as

$\rm :\longmapsto\:x = \dfrac{ - ( - 1\: - \: \sqrt{ {3}} \: i )}{2} \: or \: \dfrac{ - ( - 1\: + \: \sqrt{ {3}} \: i) }{2}$

So,

$\bf\implies \:x = - {\omega}^{2} \: \: or \: - \: \omega$

Now,

$\rm :\longmapsto\: - \omega - {\omega}^{2} = 1$

$\rm :\longmapsto\:( - \omega)( - {\omega}^{2} ) = {\omega}^{3} = 1$

Case :- 1

$\boxed{\rm :\longmapsto\:When \: x \: = \: - \: \omega}$

Consider,

$\rm :\longmapsto\: {x}^{29} + \dfrac{1}{ {x}^{89} }$

$\rm \: = \: \: {( - \omega)}^{29} + \dfrac{1}{ {( - \omega)}^{89} }$

$\rm \: = \: \: - {(\omega)}^{29} - \dfrac{1}{ {(\omega)}^{89} }$

$\rm \: = \: \: - {(\omega)}^{2} - \dfrac{1}{ {(\omega)}^{2} }$

$\rm \: = \: \: - {(\omega)}^{2} - \omega$

$\rm \: = \: \: 1$

Hence,

$\boxed{\bf :\longmapsto\: {x}^{29} + \dfrac{1}{ {x}^{89} } = 1}$

Case :- 2

$\boxed{\rm :\longmapsto\:When \: x \: = \: - \: \omega \: ^{2} }$

Consider,

$\rm :\longmapsto\: {x}^{29} + \dfrac{1}{ {x}^{89} }$

$\rm \: = \: \: {( - \omega)}^{58} + \dfrac{1}{ {( - \omega)}^{178} }$

$\rm \: = \: \: - {(\omega)}^{58} - \dfrac{1}{ {( \omega)}^{178} }$

$\rm \: = \: \: - {( \omega)}^{1} - \dfrac{1}{ {(\omega)}^{1} }$

$\rm \: = \: \: - \omega - {\omega}^{2}$

$\rm \: = \: \: 1$

Hence,

$\boxed{\bf :\longmapsto\: {x}^{29} + \dfrac{1}{ {x}^{89} } = 1}$