Polynomials (1 Viewer)

[zeek]

Hiya, how would you go about answering this:

Sketch the graphs of x²-y²=0 and xy=2 on the same axes. By observing the number of points of intersection of the graphs, deduce that the equation of z²=4i has at least one root (in fact two roots).

Now what i did is i drew the curves and they intercepted each other twice. So then i rearranged the xy=2 formula and subtracted them from each other before getting to z²=4i. Would that be a valid answer?

 

[rocky1989]

let z=x+iy, then --> z^2=4i
x^2 - y^2 + 2ixy = 4i

then you resolve real and imaginary parts...so

x^2 - y^2 = 0 and 2xy = 4

how convenient that these are the two equations that they gave us at the beginning....then you simply say that the roots are where the two graphs intersect....

the two roots you get will be (root2 + iroot2) and (-root2 -iroot2)
i think thats all you need to show

 

[Riviet]

In general, this is one of many ways to show that any complex number has two complex roots.

 

[zeek]

thx guys

 

[zeek]

ok how do you do this one P(x)=x⁴+px³+qx²+rx}s=0 has two roots which are reciprocals and two other roots which are opposites. Show that q=1+s and r=ps.

I know that you have to use the relationship between the coefficients and roots but i think im mucking up with the algebra. Can someone have a go? thx

 

[rocky1989]

well lets say that your 4 roots are: (b) , (1/b) , (h) and (-h)

then you can get the four results:

(1).........(b)+(1/b)=-p
(2).........1-(h^2)=q
(3).........-(h^2)((b) +(1/b))=-r
(4).........-(h^2)=s

from these four results it is quite easy to get the 2 answers.....maybe you had some small algebra error?

 

[zeek]

the problem i had was actually finding the value of one of the roots.. in ur working out that would be 'b'. From (1) would you expand and then use the quadratic?

 

[rocky1989]

ok, if you are trying to find the value of these roots then yes you can expand (1) and then use the quadratic formula....but unless you have some real values for this polynomial then its not going to be anything impressive.

 

[zeek]

nah there aren't any... thats it... i got square roots n stuff so i jst gave up n assumed i was wrong because it just looked to "retarded"

 

[rocky1989]

you don't need to find the value of the roots to do this question......
did you know that?

 

[zeek]

then how else would you prove that q=1+s

 

[pLuvia]

P(x)=x⁴+px³+qx²+rx+s=0
Let roots be a,1/a,b,-b
a+1/a=-p (1)
1-b²=q (2)
-b²/a=-r (3)
-b²=s (4)

Using 4 and 2
1+s=q

Using 1,3,4
-s/a=-r
a=s/r
Sub this into 1
(s/r)+(r/s)=-p
(²+r²)r=ps
Hence r=ps

 

[zeek]

Ok and how about this...
Show that cos 5@=16cos⁵@-20cos³@+5cos@. Hence solve the equation 16x⁵-20x³+5x-1=0 and deduce the exct values of cos(2pi/5) and cos(4pi/5).

I'm stuck on the second part.

 

[Riviet]

Let x=cos@ and subtract one, notice that this is identical to RHS of the result just proved.
=> cos5@-1=0

5@= 0, -2pi, 2pi, -4pi, 4pi
@=0, +2pi/5, +4pi/5

.'. x=1, cos(+2pi/5), cos(+4pi/5)

but cos is even so we have two double roots:
x=1, cos(2pi/5), cos(2pi/5), cos(4pi/5), cos(4pi/5)

Applying sums and products of roots to the quintic equation in x:

2cos(2pi/5) + 2cos(4pi/5) + 1 = 0, which becomes
cos(2pi/5) + cos(4pi/5) = -2/4 {= -b/a} (1)

cos²(2pi/5).cos²(4pi/5) = 1/16 which becomes
cos(2pi/5).cos(4pi/5)= -1/4 {=c/a} (2) since cos(4pi/5)<0 and cos(2pi/5)>0

We form a quadratic equation using the coefficients a, b and c (ax²+bx+c=0 where sum of roots = -b/a and product of roots = c/a):

4x² + 2x -1 =0

Use quadratic formula to find the exact values of cos(2pi/5) [positive] and cos(4pi/5) [negative].

I hope that's right.

 

[zeek]

sorry i meant the part where you actually find the exact values of cos(2pi/5) and cos(4pi/5).

 

[Riviet]

Updated and done.

 

[zeek]

yup! i didn't think about cos being even