Complex Roots of Polynomial (1 Viewer)

[Lukybear]

P(x) = x^4 + cx^2 + e, where c and e are real

If sqrt3 + i is a root, determine c and e, then factor into quadratic factors with linear factors

Here's my attempt:
Since c and e are real,
then sqrt3 - i is a root (complex conjugates theorem)

Also
P(-x) = P(x)
therefore P(x) is even

Hence
-(sqrt3 + i) and -(sqrt3 - i) are also roots

from there i got e = 16 (product of roots)
c= ...

However the answers have it differently.
It subbed sqrt3 + i and sqrt3 - i into P(x), then solved simultaneously acquiring:
e= - 8, c = 8

Can someone explain the flaw in my reasoning? Or perhaps answers is wrong?

 

[random-1006]

P(x) = x^4 + cx^2 + e, where c and e are real

If sqrt3 + i is a root, determine c and e, then factor into quadratic factors with linear factors

Here's my attempt:
Since c and e are real,
then sqrt3 - i is a root (complex conjugates theorem)

Also
P(-x) = P(x)
therefore P(x) is even

Hence
-(sqrt3 + i) and -(sqrt3 - i) are also roots from there i got e = 16 (product of roots)
c= ...

However the answers have it differently.
It subbed sqrt3 + i and sqrt3 - i into P(x), then solved simultaneously acquiring:
e= - 8, c = 8

Can someone explain the flaw in my reasoning? Or perhaps answers is wrong?

notice the order in which it is asked.


highlighted is wrong, im not quite sure how to explain it, but easiest way to answer questions, keep it simple!, ok from the start you have two roots and two unknowns in your polynomial, use factor theorm to get two simulataneous eqns, and solve then to get c and e, keep it simple, two unknowns and two pieces of info, factor theorm and simultaneous eqns


for second part you do a little trick

x^4 +8x^2 -8 = x^4 +8x^2 +16 -8 -16 ( complete the square on the quartic)
= (x^2 +4)^2 -24 ( difference of squares)
= (x^2 +4 +sqrt24) (x^2 +4 -sqrt24)

the question is a little strange, " quadratic factors with linear factors", confusing, that should be it

 

[Shane_]

notice the order in which it is asked.


highlighted is wrong, im not quite sure how to explain it, but easiest way to answer questions, keep it simple!, ok from the start you have two roots and two unknowns in your polynomial, use factor theorm to get two simulataneous eqns, and solve then to get c and e, keep it simple, two unknowns and two pieces of info, factor theorm and simultaneous eqns


for second part you do a little trick

x^4 +8x^2 -8 = x^4 +8x^2 +16 -8 -16 ( complete the square on the quartic)
= (x^2 +4)^2 -24 ( difference of squares)
= (x^2 +4 +sqrt24) (x^2 +4 -sqrt24)

the question is a little strange, " quadratic factors with linear factors", confusing, that should be it

Dis is pretty much rite yeah, tho wen i did it da first time, i got da same fing as you. den i checked my texbook, da guy haz brainz

 

[MetroMattums]

Do complex roots obey the same laws of symmetry?

I don't think they do :/

 

[random-1006]

Do complex roots obey the same laws of symmetry?

I don't think they do :/

im not sure, ill find out next yr lol, there is a whole subject on complex analysis in second year maths, there is some pretty interesting stuff like e^x is not periodic in the cartesian plane , but is periodic in the argand plane

 

[random-1006]

the answers are wrong.
c=-4, e=16

lol.

1. i cant be bothered checking but im pretty sure they are right, or im pretty sure yours are wrong, how do you part 2 then, you cant make a difference of squares

Edit: lol study freak retreats, its no shame, its ok to make mistakes, none of us are perfect

 

[Lukybear]

Do complex roots obey the same laws of symmetry?

I don't think they do :/

I think your right.

 

[study-freak]

lol.

1. i cant be bothered checking but im pretty sure they are right, or im pretty sure yours are wrong, how do you part 2 then, you cant make a difference of squares

Edit: lol study freak retreats, its no shame, its ok to make mistakes, none of us are perfect

gosh, i hate my internet disconnection.

i just deleted to recheck my work but nevertheless, i still think i'm right.

and you do (x-(sqrt3+i))(x-(sqrt3-i))(x+(sqrt3+i))(x+(sqrt3-i))
=(x^2-2sqrt3 x+4)(x^2+2sqrt3 x+4)

and who says you need to use difference of two squares?

 

[study-freak]

P(x) = x^4 + cx^2 + e, where c and e are real

If sqrt3 + i is a root, determine c and e, then factor into quadratic factors with linear factors

Here's my attempt:
Since c and e are real,
then sqrt3 - i is a root (complex conjugates theorem)

Also
P(-x) = P(x)
therefore P(x) is even

Hence
-(sqrt3 + i) and -(sqrt3 - i) are also roots

from there i got e = 16 (product of roots)
c= ...

However the answers have it differently.
It subbed sqrt3 + i and sqrt3 - i into P(x), then solved simultaneously acquiring:
e= - 8, c = 8

Can someone explain the flaw in my reasoning? Or perhaps answers is wrong?

This logic is correct because
(sqrt3+i)^4+c(sqrt3+i)^2+e=0
[-(sqrt3+i)]^4+c[-(sqrt3+i)]^2+e=0
thus -(sqrt3+i) is also a root.

similarly, -(sqrt3-i) is also a root.

 

[random-1006]

This logic is correct because
(sqrt3+i)^4+c(sqrt3+i)^2+e=0
[-(sqrt3+i)]^4+c[-(sqrt3+i)]^2+e=0
thus -(sqrt3+i) is also a root.

similarly, -(sqrt3-i) is also a root.

correct, but a pretty bad way of finding the unknown values

 

[Lukybear]

Yea studyfreak thxs for that. Its what i thought as well. But can there be 2 answers? Cause the book's solution is pretty sound as well.

 

[study-freak]

correct, but a pretty bad way of finding the unknown values

much simpler than actually solving eqns, i reckon
saying the expression is even suffices as a reasoning anyway.
just showing why it works.

Yea studyfreak thxs for that. Its what i thought as well. But can there be 2 answers? Cause the book's solution is pretty sound as well.

no, there's only one solution.
I did this problem in a few different ways and all gave c=-4 and e=16

 

[random-1006]

much simpler than actually solving eqns, i reckon
saying the expression is even suffices as a reasoning anyway.
just showing why it works.


no, there's only one solution.
I did this problem in a few different ways and all gave c=-4 and e=16

yeh it looks like hes correct, though i didnt think anyone would actually take the time to check the books answer as it involved so much boring expanding, the books answer looked reasonable so i used it for the second part ( however, dont forget the method i showed you, its a good method)

EDIT: huh, now that quadratic and linear factors bit in the question makes a bit of sense

 

[Trebla]

The answers appear to be wrong.



Probably would've been quicker to sub the roots and expand that ugly quartic to check but meh...

 

[Tofuu]

The answers appear to be wrong.



Probably would've been quicker to sub the roots and expand that ugly quartic to check but meh...

yea, i got unreal roots as well
least i'm not alone