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Try this! - Complex numbers and quadratic equations - JEE Main-13

If \alpha\: and \: \beta are the roots of the equation

x^{2}-x+1= 0, then \alpha ^{2009}+\beta ^{2009}=

  • Option 1)

    -2

  • Option 2)

    -1

  • Option 3)

    1

  • Option 4)

    2

 
Answers (2)
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N neha

As we have learned 

Roots of Quadratic Equation -

\alpha = \frac{-b+\sqrt{b^{2}-4ac}}{2a}

\beta = \frac{-b-\sqrt{b^{2}-4ac}}{2a}
 

 

- wherein

ax^{2}+bx+c= 0

is the equation

a,b,c\in R,\: \: a\neq 0

 

 

Cube roots of unity -

z=\left ( 1 \right )^{\frac{1}{3}}\Rightarrow z=\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3}

k=0,1,2 so z gives three roots 

\Rightarrow 1,\frac{-1}{2}+i\frac{\sqrt{3}}{2}\left ( \omega \right ),\frac{-1}{2}-i\frac{\sqrt{3}}{2}\left ( \omega^{2} \right )

- wherein

\omega=\frac{-1}{2} +\frac{i\sqrt{3}}{2},\omega^{2}=\frac{-1}{2} -\frac{i\sqrt{3}}{2},\omega^{3}=1, 1+\omega+\omega^{2}=0

1,\omega,\omega^{2} are cube roots of unity.

 

 \alpha ,\beta = \frac{1\pm \sqrt {-3}}{2}= \frac{1 \pm \sqrt{3i}}{2}= -\left ( \frac{-1\pm\sqrt{3i}}{2} \right )= -\omega ,-\omega ^2

\alpha ^{2009}+\beta ^{2009}

-\omega ^{2009}+(-\omega^2) ^{2009}

= -\omega ^2-\omega = -(\omega ^2+\omega )= -(-1)= 1

 

 

 

 


Option 1)

-2

Option 2)

-1

Option 3)

1

Option 4)

2

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