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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
avatar+95171 
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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.

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

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 Dec 10, 2018
 #4
avatar+95171 
0

I already answered you (A full day before you edited your question) 

You have not responded to my answer in any way.

What is wrong with you?  

Are you just totally rude?

 

I even read your mind and answered the question that you wanted to ask, rather than the garbled nonsense that you wrote.

 Dec 11, 2018
edited by Melody  Dec 11, 2018

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