+2729 The roots of the quadratic equation $z^2 + bz + c = 0$ are $4 + 2i$ and $-5 + 8i$. What is $b+c$?
+1950 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! :)
+1950 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! :)
+130466 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
+1950 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! :)
+1950 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
(z2−8z+20)(z2+10z+89)z4+2z3+29z2−512z+1780
So, in order for this to be valid, b would have to be z^3 and c would have to be 2z3+28z2−512z+1780
So our final answer is z4+2z3+28z2−512z+1780
Thanks! :)
