+2653 The roots of the quadratic equation $z^2 + bz + c = 0$ are $4 + 2i$ and $-5 + 8i$. What is $b+c$?
+1926 There's a problem with this question, though.
According some theorem I totally forgot,
If \(4 + 2i\) is a root, then \(4-2i\) must also be a root.
If \(-5 + 8i\) is a root, then \(-5-8i\) must also be a root.
With 4 distinct roots, then the polynomial would have degree 4, not 2,
So it is impossible to have a quadratic with these two roots.
I'm not sure if I'm mistaken, but that should be true.
Thanks! :)
+1926 There's a problem with this question, though.
According some theorem I totally forgot,
If \(4 + 2i\) is a root, then \(4-2i\) must also be a root.
If \(-5 + 8i\) is a root, then \(-5-8i\) must also be a root.
With 4 distinct roots, then the polynomial would have degree 4, not 2,
So it is impossible to have a quadratic with these two roots.
I'm not sure if I'm mistaken, but that should be true.
Thanks! :)
+129883 You are correct, NTS !!!
This can't be a quadratric because of the Conjugate Property
If a + bi is a root so is a - bi
If c + di is a root so is c - di
We would have 4 complex roots, not just two
+1926 Thanks! It's funny because I'm taking an intermediate algebra course, and this is what we are learning.
Also, NTS is such a wonderful abbreviation, LOL. Sounds quite proffesional.
Thanks! :)
+1926 So, I might have forgotten to consider than b and c could include z. I'll solve the problem like that,
Like I stated in post 1, there are 4 roots
\(4+2i, 4-2i, -5+8i, -5-8i\)
This means we have the polynomial,
\((z-4-2i)(z-4+2i)(z+5-8i)(z+5+8i)\)
Simplfying, we have
\(\left(z^2-8z+20\right)\left(z^2+10z+89\right)\\z^4+2z^3+29z^2-512z+1780\)
So, in order for this to be valid, b would have to be z^3 and c would have to be \(2z^3+28z^2-512z+1780\)
So our final answer is \(z^4+2z^3+28z^2-512z+1780\)
Thanks! :)
