SATHEE: Theory of Equations 1 Question 13

Theory of Equations 1 Question 13

14. If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0$ , then $α^{101} + β^{107}$ is equal to

(2018 Main)

(a) -1

(b) 0

(d) 2

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Answer:

Correct Answer: 14. (c)

Solution:

We have, $α, β$ are the roots of $x^{2} - x + 1 = 0$

$∵$ Roots of $x^{2} - x + 1 = 0$ are $- ω, - ω^{2}$

$\begin{aligned} ∴ Let α = - ω and β = - ω^{2} \\ \begin{aligned} \Rightarrow α^{101} + β^{107} & = (- ω)^{101} + {(- ω^{2})}^{107} = - (ω^{101} + ω^{214}) \\ = - (ω^{2} + ω) & [∵ ω^{3} = 1] \\ = - (- 1) & [∵ 1 + ω + ω^{2} = 0] \end{aligned} \\ = 1 \end{aligned}$

Theory of Equations 1 Question 13

14. If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0$ , then $α^{101} + β^{107}$ is equal to

Answer:

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Theory of Equations 1 Question 13

14. If α,β∈C are the distinct roots of the equation x2−x+1=0, then α101+β107 is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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14. If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0$ , then $α^{101} + β^{107}$ is equal to