#1
sorry for all these questions but ive recently started IB maths higher without doing Add maths for GCSE so im in need of a bit of help.

ok ive got this question:

-2+bi is a solution to z^2+az+(3+a)=0 find a and b given that they are real.

so i already know that -2-bi is another solution.... NOW WHAT!?!

Thanx in advance
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"Like a blind man at an orgy, I was going to have to feel my way through"

#2
substitute z = -2 +bi into the equation z^2 +az + (3+a) = 0. Then substitute z = -2 - bi. Then you have two equations, and two unknowns (a, and b). Expand and solve.
#3
you make something like this (-2+bi)(-2-bi) and you put them together through multiplication and distribution
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#4
i honestly dont know thats just what i would do
sorry for double post
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#5
i is the square root of negative 1

and i think the first guy got it...im doing it atm.
"Like a midget at a urinal, I was going to have to stay on my toes"

"Like a blind man at an orgy, I was going to have to feel my way through"

#6
Meh math noob time...


Is it a quadratic, having a a and c... You could use the discriminant and then factorise...Or just bash it with the full quadratic formula...Idk.
#7
Ok. So you got the other solution. Now you see that real part of both solutions is -2. Let's say that formula is (-y +- sqrt(y^2 - 4xz)/2x
x = 1
y = a
z = 3+a


Well, real part of both solutions is -2. it is equal to -y / 2x. So y = 4 x.

x = 1 therefore y = 4.

y = a so a = 4 too.

z = 3+a = 3+4=7

edit: I fotgot b XD

bi = sqrt (y^- 4 xz) / 2x

bi = sqrt (-3)

b = sqrt (3)


Just ask if you don't understand my way of solving...
Last edited by -Rodion- at Nov 13, 2008,
#9
I reeeaaally should start paying attention on math lessons :S
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#10
meh imaginary numbers are so pointless
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#12
how did u get b= sqrt3
i got (sqrt5)/2
"Like a midget at a urinal, I was going to have to stay on my toes"

"Like a blind man at an orgy, I was going to have to feel my way through"

#13
+- bi = +- sqrt (y^2 - 4xz) / 2x

y^2 - 4xz = 16 - 4*7 = -12

+- bi = sqrt (-12) / 2 = sqrt ( -12/4) = sqrt (-3)