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

2 Answers
Harish Chandra Rajpoot
Jul 28, 2018

#x^2-x+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#

P dilip_k
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#

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