Help with Question (1 Viewer)

Traveller0901 · Jan 23, 2013

Can someone please show me how to do q6 from Exercise 4.4 from the cambridge 4u textbook?

Much appeciated.

braintic · Jan 23, 2013

Attached (from a book of solutions):
View attachment Cambridge 4.4 Q6.pdf

Sy123 · Jan 23, 2013

braintic said:
Attached (from a book of solutions):
View attachment 27523

Not very elegant of a solution there, here is my own alternatively:

$x^4-x^2+1 = 0 \ \ | \times x^2$

$x^6-x^4+x^2=0 \\ \\ x^6=x^4-x^2= - 1$

$\therefore $the solutions of$ \ \ x^4-x^2+1 \ \ $ are among the solutions of$ \ \ x^6+1=0 \ \ \ $with the additional 2 solutions being created when the multiplication by x squared occured$

$\therefore $the solutions of$ \ \ P(x) \ \ $are among the values$$

$x=cis \pi /6 \ \ x=cis pi/2 \ \ x=cis5\pi/6 \ \ x=cis 7\pi/6 \ \ x= cis 3\pi/2 \ \ x = cis 11\pi/6$

From the manipulation at the start and observation, we can see that x=i, x=-i are not solutions. Hence when we factorise over R we just add the conjugates and express as quadratic factors:

$x^4-x^2+1=(x^2-2x\cos 5\pi/6 +1)(x^2-2x\cos \pi/6 + 1)$

Traveller0901 · Feb 1, 2013

@Sy123 - Thanks! I understand it now!

Help with Question (1 Viewer)

Traveller0901

Member

braintic

Well-Known Member

Sy123

This too shall pass

Traveller0901

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)