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The roots of the quadratic equation $z^2 + bz + c = 0$ are $4 + 2i$ and $-5 + 8i$. What is $b+c$?

 Jun 8, 2024

Best Answer 

 #1
avatar+759 
+1

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! :)

 Jun 8, 2024
 #1
avatar+759 
+1
Best Answer

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! :)

NotThatSmart Jun 8, 2024
 #3
avatar+129401 
0

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

 

cool cool cool

CPhill  Jun 8, 2024
 #4
avatar+759 
+1

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! :)

NotThatSmart  Jun 8, 2024
 #2
avatar+759 
+1

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! :)

 Jun 8, 2024

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