# How do I find the imaginary number solutions to this problem?

• Jan 18th 2011, 03:18 PM
thyrgle
How do I find the imaginary number solutions to this problem?
So I have the following equation:

\$\displaystyle
r^3 + 4r^2 - 24 = 0
\$

I can find the real solution of 2 by using synthetic division and finding which of the multiples of 24 ends up to have no remainder.

But, the back of the book also wants an imaginary number solution and I am lost as to how to find one. Any help would be greatly appreciated.
• Jan 18th 2011, 03:26 PM
Plato
\$\displaystyle (r-2)(r^2+6r+12)=r^3+4r^2-24\$.
Can you find the roots of \$\displaystyle r^2+6r+12~?\$
• Jan 18th 2011, 03:27 PM
e^(i*pi)
You can use the factor theorem to say that \$\displaystyle r-2\$ is a factor of \$\displaystyle f(r)\$

Hence \$\displaystyle (r-2)(Ar^2+Br+C) = r^3+4r^2-24\$ where A, B and C are constants to be found. Normally. I'd equate the coefficients but you can also do long division

edit: too slow :(
• Jan 18th 2011, 03:29 PM
Quacky
I wouldn't have even bothered with synthetic division. Factor theorem will work nicely.
You've established that r=2 is a root, therefore by the factor theorem, \$\displaystyle (r-2)\$ is a factor. So take this out (via factorization):
\$\displaystyle
(r-2)(Ar^2+Br+C)=0\$
Then just complete the square or use the formula. on the quadratic for the complex solution(s).
• Jan 18th 2011, 03:34 PM
thyrgle
@Plato:

Yeah, plug into the the quadratic equation and you get:
\$\displaystyle x = -6+/-sqrt(36 - 4(1)(12))/2(1)\$
\$\displaystyle x = -6+/-sqrt(36 - 48)/2\$
\$\displaystyle x = -6+/-sqrt(-12)/2\$
\$\displaystyle x = -6+/-2*sqrt(-3)/2\$
\$\displaystyle x = -6+/-2i*sqrt(3)/2\$
\$\displaystyle x = -3+/-i*sqrt(3)\$

Thanks!