QJEEM18Ma1d94: JEE Main 2018 Mathematics - Class 11 JEEM18Ma1d94 Solution: If are the distinct roots of the equation x 2 - x + 1 = 0, then …

Question

Verified Solution • 2025–26 Session moderate

If

\alpha ,\beta \in C

are the distinct roots of the equation
x² - x + 1 = 0, then

{\alpha ^{101}} + {\beta ^{107}}

is equal to :

Accepted Answer

Given equation,

x²

-

x + 1 = 0

Roots of this equation

x =

{{1 \pm \sqrt 3 i} \over 2}

\therefore\,\,\,

\propto \, = \,{{1 + \sqrt 3 \,i} \over 2}

and

\beta = \,{{1 - \sqrt 3 \,i} \over 2}

We know;

\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta

and

{\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto

\therefore\,\,\,

\propto \, = - {\omega ^2}

and

\beta \, = \, - \omega

\therefore\,\,\,

{ \propto ^{101}} + {\beta ^{107}}

= {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}

= {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}

= - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}

= - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)

= - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)

= - \left( {\omega + {\omega ^2}} \right)\,\,\,

[ as

\,\,\,

{\omega ^{3n}} = 1

]

= - \left( { - 1} \right)

\,\,\,\,\,\,

[as

\,\,\,

1 + \omega + {\omega ^2} = 0

]

= 1

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