how do you find the equation of a quartic with complex roots?

Question

the roots are 3+2i and 2-3i

Answer 1

Do you know how to multiply two complex numbers? if a is one complex number and b is the other complex number then in general

(x-a)(x-b) = x^2 -(a+b)x + ab

will be the quadratic equation that you are looking for. You will need to add and multiply your complex roots to get the equation.

Answer 2

I think that you have some details in the question that you left out.

First of all, a general quartic polynomial is going to have 4 roots and there is no restriction on them. They could be all different (distinct), they could be all of the same. They could be all real, all imaginary, or any possible combination.

So to answer your question, a quartic with roots 3+2i, and 2-3i, many answers are possible. The simplest one would be to make the other 2 roots zero.

Like
(x-(3+2i))*
(x-(2-3i))*
x*x

Since you never told me which roots to use, I can choose anything for the other two and zero is the easiest one.

Now, the question that YOU are trying to ask is to find a quartic with REAL RATIONAL coefficients that has 3+2i and 2-3i as its roots.

Now you are being specific and now there is only one answer.

IF we want a quartic with real rational coefficients with the given roots, then their conjugates MUST be roots of the quartic as well, otherwise the coefficients will not be real.

Which means that ALL of the roots are
3+2i
3-2i
2-3i
2+3i

Which means that the quartic is simply
(x-(3+2i))*
(x-(3-2i))*
(x-(2-3i))*
(x-(2+3i))

which multiplies to
(x^2-6x+13)(x^2-4x+13)
which gives you
x^4-10x^3+50x^2-130x+169

I hope that I multiplied that right. You might want to check that.

Answer 3

a quartic has 4 roots, but they only give you 2. The key is that they give you complex roots. If a complex number is a root, so is its conjugate:

conjugate of 3 + 2i is 3 "-" 2i
conjugate of 2 - 3i is 2 "+" 3i

Thus, the 4 roots are: 3 + 2i , 3 - 2i, 2 - 3i and 2 + 3i

To find the equation, multiply the 4 factors:
(x - [3 + 2i] )(x - [3 - 2i] )(x - [2 - 3i] )(x - [2 + 3i] )

=(x^2 - [3 + 2i]x - [3 - 2i]x + [3 + 2i] *[3 - 2i] ) * (x^2 - [2 - 3i]x - [2 + 3i]x + [2 - 3i] *[2 + 3i] )

=(x^2 - 3x - 2ix - 3x + 2ix + 9 +6i - 6i + 4) * (x^2 - 2x + 3ix - 2x - 3ix + 4 - 6i + 6i + 9)

=(x^2 - 6x + 13) * (x^2 - 4x + 13)

=x^2* (x^2 - 4x + 13) - 6x*(x^2 - 4x + 13) + 13*(x^2 - 4x + 13)
= x^4 - 4x^3 + 13x^2 - 6x^3 + 24x^2 - 78x + 13x^2 - 52x +169
= x^4 - 10x^3 + 50x^2 - 130x +169

a quartic

Answer 4

If you have a quartic equation, you have 4th power equation, i.e, x^4 power.

Given any complex root, you could find it's complex conjugate by putting a negative sign infront of the imaginary part. In your case, the other roots are:
3-2i, and 2+3i.

Therefore you have a factored equation that looks like this:

{[x-(3+2i)]*[x-(3-2i)]*[x-(2-3i)]*[x-(2+3i)]}

in otherwords:

[x-(3+2i)]*
[x-(3-2i)]*
[x-(2-3i)]*
[x-(2+3i)]

(I put it this way because it looks like the equation got cut off)
multiply it all out and that is your equation. Also note that your imaginary numbers should either cancel out or become be squared and turn into a negative one.

Answer 5

you guys are not correct. If a an equation has the roots 3+2i and 2-3i, it also has the roots 3-2i and 2+3i. You then can use product sum on the two conjugal pairs, and multiply the two quadratic equations together

Answer 6

You can build any polynomial from its roots. This is because (x-root) is always a factor of the polynomial. Yours has 2 roots--call them r1 and r2. The polynomial you want is

(x-r1)(x-r1)(x-r2)(x-r2)

Multiply this out, combine like terms, and you'r done!

Note--this is a QUARTIC (4th-order). The above poster gave you the answer for the QUADRATIC (2nd-order), so if you really meant "quadratic," you just need (x-r1)(x-r2).