# imaginary number problem

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• Sep 16th 2007, 09:20 AM
Mr_Green
imaginary number problem
List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)
• Sep 16th 2007, 09:26 AM
Jhevon
Quote:

Originally Posted by Mr_Green
List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)

here's how i'm thinking to do this. there may be an easier way:

if one root is 2 + i, then another is 2 - i, since complex roots always come in conjugates.

thus we can write: $x^4 + x^3 - 11x^2 + 9x + 20 = (x - a)(x - b)(x - (2 + i))(x - (2-i))$

where $a,b$ are the remaining 2 roots, they may be another complex conjugate pair or real (i think they are real though).

expand the right side and equate the coefficients to solve for $a$ and $b$

OR you can do it the long way and try to find real roots of the equation by plugging in factors of 20 and doing the necessary long divisions. that may work
• Sep 16th 2007, 09:39 AM
ThePerfectHacker
Quote:

Originally Posted by Mr_Green
List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)

We know that if $2+i$ is a root then $2-i$ is another one. Let $a,b$ be other roots. Then we know that $(2+i)+(2-i)+a+b = -1$ and $(2+i)(2-i)ab = 20$. Now solve.
• Sep 16th 2007, 09:41 AM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
We know that if $2+i$ is a root then $2-i$ is another one. Let $a,b$ be other roots. Then we know that $(2+i)+(2-i)+a+b = -1$ and $(2+i)(2-i)ab = 20$. Now solve.

there goes that method i saw Soroban use the other day, i still do not get how you got the answer for the sum of the roots though, i try to find that thread again to see if it was explained there

EDIT: Yup, looked at it, still don't get it

EDIT: Oh wait, i think i found something (in particular, see theorem 5): General theorems for polynomials

now to just figure out a way to memorize it
• Sep 16th 2007, 10:09 AM
Krizalid
Quote:

Originally Posted by Mr_Green
List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)

\begin{aligned}
x^4+x^3-11x^2+9x+20&=x^4+x^3-11x^2+(20x-11x)+20\\
&=x^3(x+1)-11x(x+1)+20(x+1)\\
&=(x+1)(x^3-11x+20)\\
&=(x+1)(x^3+4x^2-4x^2-16x+5x+20)\\
&=(x+1)(x+4)(x^2-4x+5)\,\blacksquare
\end{aligned}
• Sep 16th 2007, 10:42 AM
topsquark
Quote:

Originally Posted by Mr_Green
List all solutions if x^4 + x^3 - 11x^2 + 9x + 20 = 0 and if one of the roots is x = (2+i)

Another (slightly different) way:
We know that 2 + i is a solution so since the polynomial has real coefficients we know that 2 - i is also a solution. Thus
$(x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5$
is a factor of $x^4 + x^3 - 11x^2 + 9x + 20$.

Doing the division:
$x^4 + x^3 - 11x^2 + 9x + 20 = (x^2 - 4x + 5)(x^2 + 5x + 4) = 0$

Thus you can solve $x^2 + 5x + 4 = 0$ to find the other two roots.

-Dan
• Sep 16th 2007, 10:45 AM
Jhevon
Quote:

Originally Posted by topsquark
Another (slightly different) way:
We know that 2 + i is a solution so since the polynomial has real coefficients we know that 2 - i is also a solution. Thus
$(x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5$
is a factor of $x^4 + x^3 - 11x^2 + 9x + 20$.

Doing the division:
$x^4 + x^3 - 11x^2 + 9x + 20 = (x^2 - 4x + 5)(x^2 + 5x + 4) = 0$

Thus you can solve $x^2 + 5x + 4 = 0$ to find the other two roots.

-Dan

wow, there are a lot of ways to skin this cat, 5 so far. can anyone come up with another method?