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in Quadratic Equations by (114k points)
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If α and β are the roots of the equation x2 - x + 1 = 0, then α2021 + β2009 is equal to


1. -1
2. 0
3. 1
4. 2

1 Answer

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Best answer
Correct Answer - Option 3 : 1

Concept:

Consider a quadratic equation: ax2 + bx + c = 0.

Let, α and β are the roots.

  • Sum of roots = α + β = -b/a
  • Product of the roots = α × β = c/a

 

Cube Roots of unity are 1, ω and ω2

Here, ω = \(\frac{{ - {\rm{\;}}1{\rm{\;}} + {\rm{\;\;i}}\sqrt 3 }}{2}\)  and ω2 = \(\frac{{ - {\rm{\;}}1{\rm{\;}} - {\rm{\;\;i}}\sqrt 3 }}{2}\)

Property of cube roots of unity

  • ω3 = 1
  • 1 + ω + ω2 = 0
  • ω = 1 / ω 2 and ω2 = 1 / ω
  • ω3n = 1

 

Calculation:

Here, α and β are the roots of the equation x2 - x + 1 = 0

Sum of roots = α + β = 1

Product of roots = α × β = 1

So, α = \(\frac{{ {\rm{\;}}1{\rm{\;}} + {\rm{\;\;i}}\sqrt 3 }}{2}\) = -ω2 and β = \(\frac{{ {\rm{\;}}1{\rm{\;}} - {\rm{\;\;i}}\sqrt 3 }}{2}\) = -ω

Now, α3 = (-ω2)3 = -ω6 = -1

β3 = (-ω)3 = -ω3 = -1

 

α2021 + β2009 = \(\rm \frac{\alpha^{2022} }{\alpha}\rm +\frac{\beta ^{2010}}{\beta}\)

\(\rm \frac{(\alpha^3)^{674} }{\alpha}\rm +\frac{(\beta ^3)^{670}}{\beta}\)

\(\rm \frac{(-1)^{674} }{\alpha}\rm +\frac{(-1)^{670}}{\beta}\) 

\(\rm \frac{1 }{\alpha}\rm +\frac{1}{\beta}=\frac{\alpha+\beta }{\alpha\beta}\)

= 1                           

Hence, option (3) is correct. 

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