Question

Find all distinct roots(real or complex) of
z^2 + (9+3i)z + (24+11i) in the form a+bi.
Hint: You may have to complete the square

Answer 1

http://answers.yahoo.com/question/index?qid=20121124223053AAchfPc
the bottom explanation of this link might be it, but i dont understand it

Answer 2

have you figured anything out?

Answer 3

\[\large z^2+(9+3i)z+(24-11i)=0\]\[\large z^2+(9+3i)z=-(24-11i)\]to complete squares, add (b/2)^2 to both sides, where b=9+3i \[\large z^2+(9+3i)z+\left( \frac{ 9+3i }{ 2 } \right)^2=-(24-11i)+\left( \frac{ 9+3i }{ 2 } \right)^2\]

Answer 4

yeah im at this step... what do you do next?

Answer 5

\[(z + (\frac{ 9 + 3i }{ 2 }))^2 = -24 + 11i + 18 +\frac{ 27i }{ 2 }\]

Answer 6

\[z + (\frac{ 9+3i }{ 2 }) = \pm \sqrt{\frac{ -12 + 5i}{ 2 }} \]

Answer 7

i dunno, lol

Answer 8

i dont get how the person completed the square in the yahoo link..

Answer 9

\[z = -(\frac{ 9+3i }{ 2}) \pm \sqrt{\frac{ -12 + 5i }{ 2 }}\]

Answer 10

it would be so much easier if the square root wasnt there...we could just add them up easily

Answer 11

from the yahoo answers link:
z² + (2−4i)z = (11+10i)

z² + (2−4i)z + (-3 -4i) = (11+10i) + (-3 -4i)

Answer 12

what i dont get is where the (-3-4i) comes from