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Let $a$ and $b$ be real numbers. The complex number $4 - 5i$ is a root of the quadratic $z^2 + (a + 8i) z + (-39 + bi) = 0.$What is the ot

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Let $$a$$ and $$b$$ be real numbers. The complex number $$4 - 5i$$ is a root of the quadratic

$$z^2 + (a + 8i) z + (-39 + bi) = 0$$
What is the other root?

Edit: Sorry wrong equation

Dec 9, 2018
edited by RektTheNoob  Dec 11, 2018

#3
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There seems to be 2 different questions here.

I will answer the first one.

Let  a  and  b  be real numbers. The complex number    4 - 5i     is a root of the quadratic

$$z^2 + (a + 8i) z + (-39 + bi) = 0$$.

I am told that one solution is z=4-5i

so

$$(4-5i)^2 + (a + 8i) (4-5i) + (-39 + bi) = 0\\ \text{expand and simplify and we get}\\ (4a-8)+(b-5a-8)i=0\\ so\\ 4a-8=0\\ a=2\\ b-5a-8=0\\ b=18\\ \text{so the equation becomes}\\ z^2 + (2 + 8i) z+ (-39 + 18i) = 0\\$$

Now I will use the quadratic formula to solve this.

$$z^2 + (2 + 8i) z+ (-39 + 18i) = 0\\ z = {-(2 + 8i) \pm \sqrt{\triangle} \over 2}\\ (-1-4i)\pm\frac{\sqrt{\triangle}}{2}=4-5i \qquad \text{I'm not sure if it is + or - so i will keep it together}\\ \pm\frac{\sqrt{\triangle}}{2}=5-i\\ \text{So the solutions are}\\ z=-1-4i\pm (5-i)\\ z=-1-4i+ (5-i) \qquad z=-1-4i- (5-i)\\ z=4-5i \qquad \qquad \quad \qquad z=-6-3i\\~\\ \text{So the other root is } -6-3i$$

Dec 10, 2018
#4
+110191
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You have not responded to my answer in any way.

What is wrong with you?

Are you just totally rude?