Unity Application

Algebra Level 4

Let α , β , γ \alpha, \beta, \gamma be the roots of the equaiton z 3 = 1 z^3= 1 such that β \beta is in the form a + i b c \frac { -a + i \sqrt b }{c} where a , b , c a,b,c are positive integers and γ \gamma as the conjugate of β \beta .

Given that x = β γ α 1 x = \beta - \gamma - \alpha - 1 . Find the value of x 4 + 3 x 3 + 2 x 2 11 x 6 x^4 + 3x^3 + 2x^2 - 11x - 6 .


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Clearly α \alpha , β \beta and γ \gamma are the cubic roots of unity. We know that α = 1 \alpha=1 , β = e 2 π i 3 \beta=e^\dfrac{2\pi i}{3} and γ = e 2 π i 3 \gamma=e^{-\dfrac{2\pi i}{3}} . So,

x = e 2 π i 3 e 2 π i 3 1 1 x = 2 i sin ( 2 π 3 ) 2 x = 3 i 2 x=e^\dfrac{2\pi i}{3}-e^{-\dfrac{2\pi i}{3}}-1-1 \\ x=2i \sin \left(\dfrac{2 \pi}{3}\right)-2 \\ x=\sqrt{3}i-2

Manipulate that expression a little:

x + 2 = 3 i ( x + 2 ) 2 = ( 3 i ) 2 x 2 + 4 x + 4 = 3 x 2 + 4 x + 7 = 0 x+2=\sqrt{3}i \\ (x+2)^2=(\sqrt{3}i)^2 \\ x^2+4x+4=-3 \\ x^2+4x+7=0

Now, using long polynomial division we know that:

x 4 + 3 x 3 + 2 x 2 11 x 6 = ( x 2 + 4 x + 7 ) ( x 2 x 1 ) + 1 x 4 + 3 x 3 + 2 x 2 11 x 6 = ( 0 ) ( x 2 x 1 ) + 1 x 4 + 3 x 3 + 2 x 2 11 x 6 = 1 x^4+3x^3+2x^2-11x-6=(x^2+4x+7)(x^2-x-1)+1 \\ x^4+3x^3+2x^2-11x-6=(0)(x^2-x-1)+1 \\ x^4+3x^3+2x^2-11x-6=\boxed{1}

Nicer way of doing it than me lol... I literally chugged all through the polynomial and evaluated it. Great job!

Kevin Mano - 5 years, 10 months ago

Up voted. Nice approach.

Niranjan Khanderia - 4 years, 10 months ago

(\beta,\gema=\dfrac{-1\pm\sqrt3} 2. \alpha=1.\
\therefore\ x=-2+\sqrt3i.\
x= - 2 + \sqrt3i.\
x^2= 1 - 4\sqrt3i.\
x^3= 10 + 9\sqrt3i.\
x^4= - 47 - 8\sqrt3i.\
\therefore x^4+3x^3+2x^2 - 11x - 6=1.



0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...