Complex Roots of Polynomial (2 Viewers)

Lukybear · Jun 26, 2010

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 · Jun 26, 2010

Lukybear said:
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_ · Jun 26, 2010

random-1006 said:
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 · Jun 26, 2010

Do complex roots obey the same laws of symmetry?

I don't think they do :/

random-1006 · Jun 26, 2010

MetroMattums said:
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 · Jun 26, 2010

study-freak said:
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 · Jun 26, 2010

MetroMattums said:
Do complex roots obey the same laws of symmetry?

I don't think they do :/

I think your right.

study-freak · Jun 26, 2010

random-1006 said:
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 · Jun 26, 2010

Lukybear said:
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 · Jun 26, 2010

study-freak said:
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 · Jun 26, 2010

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 · Jun 26, 2010

random-1006 said:
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.

Lukybear said:
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 · Jun 26, 2010

study-freak said:
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 · Jun 26, 2010

The answers appear to be wrong.

$roots are $\sqrt{3}+i,\sqrt{3}-i,\alpha,\beta\\\\$Sum of roots:\\$ 2\sqrt{3}+\alpha+\beta=0\\\Rightarrow \alpha + \beta = -2\sqrt{3}\\\\$Sum of roots in triplets:$\\(\sqrt{3}+i)(\sqrt{3}-i)\alpha + (\sqrt{3}+i)(\sqrt{3}-i)\beta + (\sqrt{3}+i)\alpha\beta+(\sqrt{3}-i)\alpha\beta=0\\\Rightarrow 4(\alpha+\beta)+2\sqrt{3}\alpha\beta=0\\\Rightarrow \alpha\beta=4\\\\$Therefore a quadratic equation with roots $\alpha $ and $\beta$ is $:\\x^2+2\sqrt{3}x+4=0\\\Delta=-4<0\\\therefore \alpha,\beta $ are non-real, hence all roots of $P(x)$ are non-real.\\\\If $c=8,e=-8$ as suggested by the answers$\\P(x)=x^4+8x^2-8\\=x^4+8x^2+16-24\\=(x^2+4)^2-24\\=(x^2+4-\sqrt{24})(x^2+4+\sqrt{24})\\$Solve $ P(x)=0\\x^2=\sqrt{24}-4>0\\$OR $\\x^2=-\sqrt{24}-4<0\\\\$The first condition implies that a pair of roots are REAL which contradicts the fact we found that none of the reals should be real from above.$

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

Tofuu · Jun 27, 2010

Trebla said:
The answers appear to be wrong.

$roots are $\sqrt{3}+i,\sqrt{3}-i,\alpha,\beta\\\\$Sum of roots:\\$ 2\sqrt{3}+\alpha+\beta=0\\\Rightarrow \alpha + \beta = -2\sqrt{3}\\\\$Sum of roots in triplets:$\\(\sqrt{3}+i)(\sqrt{3}-i)\alpha + (\sqrt{3}+i)(\sqrt{3}-i)\beta + (\sqrt{3}+i)\alpha\beta+(\sqrt{3}-i)\alpha\beta=0\\\Rightarrow 4(\alpha+\beta)+2\sqrt{3}\alpha\beta=0\\\Rightarrow \alpha\beta=4\\\\$Therefore a quadratic equation with roots $\alpha $ and $\beta$ is $:\\x^2+2\sqrt{3}x+4=0\\\Delta=-4<0\\\therefore \alpha,\beta $ are non-real, hence all roots of $P(x)$ are non-real.\\\\If $c=8,e=-8$ as suggested by the answers$\\P(x)=x^4+8x^2-8\\=x^4+8x^2+16-24\\=(x^2+4)^2-24\\=(x^2+4-\sqrt{24})(x^2+4+\sqrt{24})\\$Solve $ P(x)=0\\x^2=\sqrt{24}-4>0\\$OR $\\x^2=-\sqrt{24}-4<0\\\\$The first condition implies that a pair of roots are REAL which contradicts the fact we found that none of the reals should be real from above.$

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

Complex Roots of Polynomial (2 Viewers)

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