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marigirl
 one year ago
I have made a start, I am a little stuck in the middle:
w = 2 – 3i is a solution of the equation
\[ 3w^314 w^2+Aw– 26 = 0\]
where A is real. Find the value of A and the other two solutions of the equation
marigirl
 one year ago
I have made a start, I am a little stuck in the middle: w = 2 – 3i is a solution of the equation \[ 3w^314 w^2+Aw– 26 = 0\] where A is real. Find the value of A and the other two solutions of the equation

This Question is Closed

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2I know the solution has to be a conjugate pair, so I did (w(2+3i) )(w(23i) and got \[w^24w+13 \] a bit stuck from here............plz help :)

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2OMGGGG I SOLVED IT!!!!!!!!!!!

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2:D :D Thank guys for dropping by and checking out the question!

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2If anyone is interested in knowing.. I mirrored w^24w+13 to look like my original cubic by multiplying it with (kw+c) then matched up the coefficients to the values i have and then found the values i didnt have (i.e A)

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2@LynFran thanks for the medal :D :D :D

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you could also have done long division and set A so that the remainder is zero
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