Polynomial equation w/ all complex roots?

Question

What's a polynomial equation with all complex roots? I don't quite understand what a coplex root is.

Answer 1

complex roots are the roots that are in the form of this:
sqrt( - x). Meaning that you are taking a square root of a negative number. As you probably know we do this by putting a i. such as the sqrt(-2) is just sqrt(2) * i. We cant take sqrt(-2) but we take out the negative by multiplying by i. So it would be sqrt(2)*i.

I guess you could write any polynomial equation. The question you mentioned is too broad. Maybe you meant that "write a polynomial equation with complex roots".
In that case it would be just
X^2 + x + 1 = 0
When you use the quadratic formula here you will notice that the number under the radical sign is negative.

It would be sqrt(-3) and just take out the - by multiplying by i
so, it's sqrt(3) * i

Answer 2

Hopefully you know what a complex number is.

A complex root is a root that is not a real number.

For example, there are two roots to x^2+1, namely i and -i. Neither is real. Both are considered complex.

There are also two complex roots to x^2+4 = 0. So there are four complex roots, and no real ones, to (x^2+1)(x^2+4)=0.

Or multiply out (x+1+i)(x+1-i). You'll see that it's a real polynomial, and obviously it has only complex roots. (It's x^2+2x+2).

Or do the same thing for (x + 3+5i)(x +3-5i).

Perhaps you even start to notice a pattern!

One tip -- ALL polynomials over the reals of ODD order have at least one real root. The graphs always go from -infinity to +infinity or vice versa, and hence cross the x-axis. Only polynomials of EVEN order can be complex-roots-only.

Another tip, as per my examples above -- if a+bi is a root of a real polynomial, so is its CONJUGATE a-bi. That's because conjugation passes through addition, multiplication, and hence the whole polynomial function, and because the conjugate of 0 is of course just 0.

Answer 3

If the roots are complex then
b^2-4ac<0 or b^2<4ac where a,b and c are the coefficients of the eq in the standard form
ax^2+bx+c The reason the root is complex is that it has a real part: -b/2a and a non zero imaginary part: +/- i {sqrt[4ac-b^2]}/2a

Answer 4

because of the fact the coefficients are genuine, any complicated roots are conjugates. So the 1st one, call it f(x), has roots x = a million/2, 2-3i, 2+3i, and the 2d, call it g(x) has roots x = 2i, -2i, -2/3, 3. Use Vieta's kinfolk (in the link under) to discover the cofficients of f(x) and g(x). EDIT: @ Mathmom, or irrational! :)