Question

Determine the number and type of complex solutions and possible real solutions for each of the following equations.
2) 4x^3 – 12x + 9 = 0
@mathmale @michele_laino

Answer 1

Would I start by doing b^2 -4ac?

Answer 2

no, since it is a third degree equation

Answer 3

ok so what would i do?

Answer 4

I'm thinking...

Answer 5

can you tell us how many solutions you will have? There is a theorem you will have learned that will tell you this.

Answer 6

Well, you have a few options.
1) You can plug it into your calculator and find how many zeros there are.
2) You can use the p/q method.

There should be a few more

Answer 7

Wouldnt there be 3 solutions since it is to the third degree?

Answer 8

yes! such equations has three solutions

Answer 9

equation*

Answer 10

it is suffice to apply the fundamental theorem of algebra

Answer 11

ok so what do i do to determine the number and type of complex solutions and possible real solutions for each of the following equations.

Answer 12

I have to be able to show work

Answer 13

So, the FTA tells us how many roots there are. First thing we want to do for work is try and factor this into something nice.

Answer 14

so let's try and reduce this to some sort of multiplication we already know how to find roots of.

Answer 15

Ok how woild we do that

Answer 16

play with it algebraically. should be able to get (x+__)(quadratic)=0

Answer 17

I honestly have no idea

Answer 18

but check your op values first because the numbers for this are really really nasty

Answer 19

for your reference of how nasty they are
http://www.wolframalpha.com/input/?i=+4%28x%5E3+ –+3x%29+%2B+9+%3D+0

Answer 20

oh wait, we don't actually have to find the roots just say what they are?

Answer 21

Determine the number and type of complex solutions and possible real solutions for each of the following equations.

Answer 22

so it doesnt say we have to find them

Answer 23

I think it is better to go into the complex plane \(z\), namely, I rewrite your equation, like below:
\[\Large 4{z^3} - 12z + 9 = 0,\quad z = x + iy\]

Answer 24

so, by FTA how many solutions do we have to have?

Answer 25

then, when you graph it in the real plane, how many zeroes are there?

Answer 26

if only 1, we know that there must be a complex root and those only come in pairs.

Answer 27

I think that'd be my approach.

Answer 28

is the 0 where x= 0

Answer 29

descarte's rule of signs tells us there is exactly one negative real zero

Answer 30

if I multiply both sides by the cnjugate \({\bar z}\), I get:
\[\Large 4{z^2}{\left| z \right|^2} - 12{\left| z \right|^2} + 9\bar z = 0\]

Answer 31

conjugate*

Answer 32

then using calculus one can show that the negative root is the only real one...thus the rest are complex

Answer 33

Ok but somehow i have to show my work with an equation

Answer 34

So how ould i do that

Answer 35

next I replace these quantities:
\[\Large \begin{gathered}
{z^2} = {x^2} - {y^2} + 2ixy \hfill \\
{\left| z \right|^2} = {x^2} + {y^2} \hfill \\
\bar z = x - iy \hfill \\
\end{gathered} \]

Answer 36

We have never used the z one either

Answer 37

@mathmale

Answer 38

?

Answer 39

sorry, my procedure, doesn't work

Answer 40

do you know any other way

Answer 41

I'm searching for it, please wait...

Answer 42

@mathmale

Answer 43

we can compute the real solution easily, if we solve graphically, this algebraic system:
\[\left\{ \begin{gathered}
y = 4{x^3} \hfill \\
y = 12x - 9 \hfill \\
\end{gathered} \right.\]
here is the solution:

Answer 44

I dont think that is the real solutionis it?

Answer 45

as we can see, we have only one real solution

Answer 46

There has to be one way where we can find b0oth real and complex i just dont know what that equation would be

Answer 47

@mathmale do you have any idea

Answer 48

actually the real solution is right but how do i show how to find the complex>?

Answer 49

?

Answer 50

@triciaal

Answer 51

sorry