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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Complex Number

BITSAT 2013 · Mathematics · Complex Number

If α and β are the roots of x^ 2 -x+1=0, then the equation whose...

If α\alpha and β\beta are the roots of x2−x+1=0,x^{2}-x+1=0, then the equation whose roots are α100\alpha^{100} and β100\beta^{100} are

Options

  1. Ax2−x+1=0x^{2}-x+1=0
  2. Bx2+x−1=0x^{2}+x-1=0
  3. Cx2−x−1=0x^{2}-x-1=0
  4. Dx2+x+1=0x^{2}+x+1=0

Correct answer

D. x2+x+1=0x^{2}+x+1=0

Step-by-step solution

We have x3+1≡(x+1)(x2−x+1)x^{3}+1 \equiv(x+1) \left(x^{2}-x+1\right). Therefore, \alpha and \beta are the complex cube roots of -1 so that we may take α=−ωandβ=−ω2,\alpha=-\omega and \beta=-\omega^{2}, where ω≠1\omega \neq1 is a cube root of unity. Thus α100=(−ω)100=ωandβ100=(−ω2)100=ω2,\alpha^{100}=(-\omega)^{100}=\omega and \beta^{100}=\left(-\omega^{2}\right)^{100}=\omega^{2}, so that the required equation is \mathrmx2+x+1=0\mathrmx^{2}+ \mathrm{x}+1=0
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