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# Algebra

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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

#1
+759
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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
+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! :)

NotThatSmart Jun 8, 2024
#3
+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

CPhill  Jun 8, 2024
#4
+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
+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