msmarvelous
New member
- Joined
- Aug 10, 2009
- Messages
- 15
In general, what kind of complex roots always occur as conjugates?
Thank you
Thank you
Conjugates
- Thread starter msmarvelous
- Start date
To give you an understanding of Complex conjugate roots here are a few worked out so you can compare the base equations with the resulting roots.
1. Finding the roots for \(\displaystyle x^2+x+4\)
using the quadradic formula \(\displaystyle \frac{-b\pm \sqrt(b^2-4ac)}{2a}\) because the equation is in the form\(\displaystyle ax^2+bx+c\)
a=1 b=1 c=4
with a,b,and c plugged in we get \(\displaystyle \frac{-1\pm \sqrt(1^2-4(1)(4))}{(2)(1)}\)
Which finally yields \(\displaystyle -\frac{1}{2}\pm 2i\) which are complex conjugate roots \(\displaystyle -\frac{1}{2} + 2i\) and \(\displaystyle -\frac{1}{2} - 2i\)
2. lets do a larger polynomial of \(\displaystyle 4x^3+3x^2+3x+2\) This Polynomial came be broken up into \(\displaystyle (x+1)(4x^2+3x+2)\)
The the first term gives us one of our roots \(\displaystyle x=-1\) but the one of concern is the second one.
Using the quadratic formula again with a=4 b=3 and c=2
\(\displaystyle \frac{-3\pm \sqrt(3^2-4(4)(2))}{(2)(4)}\)
Simplifying we get another complex conjugate of \(\displaystyle -\frac{3}{8}\pm \frac{1}{8}i\sqrt(23)\)
which are complex conjugate roots \(\displaystyle -\frac{3}{8} + \frac{1}{8}i\sqrt(23)\) and \(\displaystyle -\frac{3}{8} - \frac{1}{8}i\sqrt(23)\)
3. to give some contrast here is a polynomial with real roots. \(\displaystyle x^2+8x-4\)
the quadratic formula yields \(\displaystyle -4\pm2 \sqrt(5)\) from this are roots are real and distinct \(\displaystyle 0.47\) and \(\displaystyle -8.47\)
1. Finding the roots for \(\displaystyle x^2+x+4\)
using the quadradic formula \(\displaystyle \frac{-b\pm \sqrt(b^2-4ac)}{2a}\) because the equation is in the form\(\displaystyle ax^2+bx+c\)
a=1 b=1 c=4
with a,b,and c plugged in we get \(\displaystyle \frac{-1\pm \sqrt(1^2-4(1)(4))}{(2)(1)}\)
Which finally yields \(\displaystyle -\frac{1}{2}\pm 2i\) which are complex conjugate roots \(\displaystyle -\frac{1}{2} + 2i\) and \(\displaystyle -\frac{1}{2} - 2i\)
2. lets do a larger polynomial of \(\displaystyle 4x^3+3x^2+3x+2\) This Polynomial came be broken up into \(\displaystyle (x+1)(4x^2+3x+2)\)
The the first term gives us one of our roots \(\displaystyle x=-1\) but the one of concern is the second one.
Using the quadratic formula again with a=4 b=3 and c=2
\(\displaystyle \frac{-3\pm \sqrt(3^2-4(4)(2))}{(2)(4)}\)
Simplifying we get another complex conjugate of \(\displaystyle -\frac{3}{8}\pm \frac{1}{8}i\sqrt(23)\)
which are complex conjugate roots \(\displaystyle -\frac{3}{8} + \frac{1}{8}i\sqrt(23)\) and \(\displaystyle -\frac{3}{8} - \frac{1}{8}i\sqrt(23)\)
3. to give some contrast here is a polynomial with real roots. \(\displaystyle x^2+8x-4\)
the quadratic formula yields \(\displaystyle -4\pm2 \sqrt(5)\) from this are roots are real and distinct \(\displaystyle 0.47\) and \(\displaystyle -8.47\)