Question

PLEASE HELP ASAP!!!! Give one example of how it is possible for a 9th degree function to have 2 real solutions and 4 imaginary.​

Answer 1

You may have used the quadratic formula before, but this time we are working with quadratic equations with two imaginary solutions. All this means is that there are negative numbers under the radical that have to be converted into imaginary numbers.

Answer 2

It is not possible for a 9th degree polynomial to have exactly 2 real solutions and 4 imaginary solutions.

By the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n complex roots, counting multiplicities. Since 2 + 4i is a complex number, it can only be a root of a polynomial of degree 1 (a linear function) or higher.

If a 9th degree polynomial has 2 real roots, then it must have 7 complex roots (counting multiplicities). These complex roots can be either purely imaginary or have non-zero real and imaginary parts. However, it cannot have exactly 4 purely imaginary roots and 3 with non-zero real and imaginary parts, since complex roots of a polynomial occur in conjugate pairs. Therefore, it is not possible for a 9th degree polynomial to have exactly 2 real solutions and 4 imaginary solutions.

Explanation: