Question

Show me how to get the answer please:

Find all distinct roots (real or complex) of z^2+(5+7i)z+(−4+19i) in form of a+bi

Answer 1

\[z=\frac{-5-7i \pm \sqrt{(5+7i)^2-4(1)(-4+19i)}}{2}\]
\[z=\frac{-5-7i \pm \sqrt{25+70i-49+16-76i}}{2}\]

Answer 2

But there must be distinct root answer. Should the complex answers cancel out to get the overall form of a+bi?

Answer 3

I don't believe they can.

Answer 4

That is frustrating. Because on my software it says it has an answer

Answer 5

\[z=\frac{-5-7i \pm \sqrt{-8-6i}}{2}\]

Answer 6

Based from that I cannot reduce it down due to that -6i.

Answer 7

I think it will factor.

Answer 8

(x+(2+5i))(x+(3+2i))=0

Answer 9

could i just take the sq root of the 8 and the 6 and i individually?

Answer 10

no

Answer 11

can you show me how you got that factor?

Answer 12

Trial and error. I focused on how to get the 19i

Answer 13

Also, I think if you work on it you will find that the radicand -8-6i is a perfect square.

Answer 14

If you had (1-3i)(1-3i)
the first would be 1 and the last would be 9i^2 which is -9 and 1 - 9 = -8
Now...the outer is -3i and the inner is -3i so that is -6i and so that works.

Answer 15

\[z=\frac{-5-7i \pm (1-3i)}{2}\]

Answer 16

hm, interesting. Well I will keep at it for sure

Answer 17

\[z=\frac{-4-10i}{2}\]

Answer 18

\[z=\frac{-6-4i}{2}\]

Answer 19

thanks

Answer 20

yw