#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

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

#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

#4

i honestly dont know thats just what i would do

sorry for double post

sorry for double post

#5

i is the square root of negative 1

and i think the first guy got it...im doing it atm.

and i think the first guy got it...im doing it atm.

#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.

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...

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,*

#8

yeah, I get a = 4, and b = sqrt(3)

#9

I reeeaaally should start paying attention on math lessons :S

#10

meh imaginary numbers are so pointless

#11

bi - 2 = lonely man

#12

how did u get b= sqrt3

i got (sqrt5)/2

i got (sqrt5)/2

#13

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

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

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

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

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

#14

So when does this kind of stuff fit into your daily life ?