# roots of polynomial equations (complex #s)

• May 6th 2008, 08:07 PM
phthiriasis
roots of polynomial equations (complex #s)
so im just starting to learn about roots of polynomial equations with roots that are complex numbers and i dont understand them at all.. i hope you can help

A. Use the quadratic formula to find the roots of the equation x^2+4x+5=0. Simplify and compare the roots. what do you notice?
B. write a quadratic equation with integral coefficients such that one of its roots is 4-5i
C. Write a quartic equation with integral coefficients and with roots 7i and -3i
.....
so for A i think its x= -2+i and x=-2-i but i dont "notice" anything and have no idea how to do the other questions.
im thinking its something simple and im going to feel stupid once i find out the answer -.-
• May 6th 2008, 08:23 PM
Solan
The roots are conjugate pairs
• May 6th 2008, 08:25 PM
TheEmptySet
Quote:

Originally Posted by phthiriasis
so im just starting to learn about roots of polynomial equations with roots that are complex numbers and i dont understand them at all.. i hope you can help

A. Use the quadratic formula to find the roots of the equation x^2+4x+5=0. Simplify and compare the roots. what do you notice?
B. write a quadratic equation with integral coefficients such that one of its roots is 4-5i
C. Write a quartic equation with integral coefficients and with roots 7i and -3i
.....
so for A i get x= -2+i and x=-2-i but i dont "notice" anything and have no idea how to do the other questions.
im thinking its something simple and im going to feel stupid once i find out the answer -.-

What they wanted you to notice is that $-2+i \mbox{ and } -2-i$

are conjugates two complex numbers are conjugate if

$a+bi \mbox{ and } a-bi$ the numbers are the same, but the immaginary parts have opposite signs.

Now with this new knowlege(Wink)

we can find the roots of the other quadratic.

since we know that 4-5i is a root then 4+5i must also be a root.

If c is a root of a polynomial then x-c is a facor of the polynomial

now comes the fun part

We multiply out our two factors.

$[x-(4-5i)][x-(4+5i)]=x^2-(4+5i)x-(4-5i)x+(4-5i)(4+5i)=$
$x^2-4x-5ix-4x+5ix+16+20i-20i-25i^2=x^2-8x+16-25(-1)=$
$x^2-8x+41$

I hope this helps
see what you can do with the next one.
(Rock)
• May 6th 2008, 08:48 PM
phthiriasis
thanks a lot TheEmptySet
so i tried the next one..
since 7i and -3i are roots the other roots are -7i and 3i
so i multiplied it all out..
(x-7i)(x+7i)(x-3i)(x+3i) and ended up with
x^4+58x^2+441
is this right?^^
thanks again!