If α,β are the roots of x²-x+1=0,then the quadratic equation whose roots are α²⁰¹⁵,β²⁰¹⁵ is ?

2 Answers

x^2-x+1=0x2x+1=0

Explanation:

The roots \alpha\ \ &\ \ \beta of quadratic equation: x^2-x+1=0 are given as

\alpha, \beta=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(1)}}{2(1)}

=\frac{1\pmi\sqrt{3}}{2}

Let \alpha=\frac{1+i\sqrt{3}}{2}

\alpha=e^{i\pi/3}

\therefore \alpha^{2015}=(e^{i\pi/3})^{2015}

=e^{i{2015\pi}/3}

=\cos({2015\pi}/3)+i\sin({2015\pi}/3)

=\cos(\pi/3)-i\sin(\pi/3)

=\frac{1-i\sqrt{3}}{2}

Let beta=\frac{1-i\sqrt{3}}{2}

\beta=e^{-i\pi/3}

\therefore \beta^{2015}=(e^{-i\pi/3})^{2015}

=e^{-i{2015\pi}/3}

=\cos(-{2015\pi}/3)+i\sin(-{2015\pi}/3)

=\cos(\pi/3)+i\sin(\pi/3)

=\frac{1+i\sqrt{3}}{2}

Since, the roots of new quadratic equation are same as that of given equation x^2-x+1=0 hence new quadratic equation

x^2-x+1=0

Jul 28, 2018

If alpha and beta are two different roots of the quadratic equation x^2-x+1=0 then

alpha+beta=1and alphabeta=1

Also
alpha^2-alpha+1=0 and beta^2-beta+1=0

So
alpha^3=alpha^3+1-1=(alpha+1)(alpha^2-alpha+1)-1
=(alpha+1)xx0-1=-1

Similarly beta^3=-1

Now alpha^2015=alpha^(3xx671+2)=(alpha^3)^671 alpha^2=(-1)^871alpha^2=-alpha^2

similarly beta^2015=-beta^2

So sum of the roots of new equation

alpha^2015+beta^2015

=-alpha^2-beta^2

=-(alpha^2+beta^2)

=-((alpha+beta)^2-2alphabeta)=-(1^2-2*1)=1

And the product of the roots

=alpha^2015beta^2015=(alphabeta)^2015=1^2015=1

Nowthe quadratic equation whose roots are alpha^2015andbeta^2015 will be

x^2-("sum of the roots")x+"product of the roots"=0

=>x^2-x+1=0