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Let \alpha and \beta be two roots of the equations x^2 + 2x + 2 = 0, then \alpha^{15} + \beta^{15} is equal to:

  • Option 1)

    -256

  • Option 2)

    512

  • Option 3)

    -512

  • Option 4)

    256

Answers (1)

best_answer

 

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

 

 

Polar Form of a Complex Number -

z=r(cos\theta+isin\theta)

- wherein

r= modulus of z and \theta is the argument of z

roots of quadratic eq. x^{2}+2x+2=0   are -1\pm i

let \alpha =-1+ i  and  \beta =-1-i

\alpha ^{15}+\beta ^{15}=\left ( -1+i \right )^{15}+\left ( -1-i \right )^{15}

                     =\left ( \sqrt{2} \right )^{15}\left [ \left ( e^{i\frac{3\pi }{4}} \right )^{15}+\left ( e^{i\frac{-3\pi }{4}} \right )^{15} \right ]

\because e^{i0}=\cos \left ( \theta \right )+i\sin \left ( \theta \right )

\alpha ^{15}+\beta ^{15}=\left ( \sqrt{2} \right )^{15}\left [ 2\cos \left ( 11\pi +\frac{\pi }{4} \right ) \right ]

                     =\left ( \sqrt{2} \right )^{15}\left (- \frac{1}{\sqrt{2}} \right )\times 2

                    =-256

 


Option 1)

-256

Option 2)

512

Option 3)

-512

Option 4)

256

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