Is it possible for the solution set of a quadratic equation with real coefficients to consist of one real number and one nonreal number?

Answer 1

Complex Conjugate Root Theorem

By knowing the Complex Conjugate Root Theorem, we know that all imaginary roots in the form a + bi have a conjugate a - bi which is also a root. Thus, there are either 2 or 0 imaginary roots to the equation and it would not be possible to have only one non-real number.

The Discriminant

The discriminant of any quadratic equation in standard form is b2 - 4ac. If the discriminant is a positive number, the equation has 2 real roots. If the discriminant equals 0, there is 1 real root. If the discriminant is negative, there are 2 imaginary roots. Thus it is not possible to have one real and one non-real root to a quadratic equation with real coefficients.

Proof By Contradiction

Assume there exists a polynomial of degree 2 with real coefficients with one real solution x= R where R is a real number and one non-real solution x = a + bi where a and b are real numbers and i is defined as the square root of -1. This quadratic equation can now be written as:

(x - R)(x - (a + bi)) = x2 - Rx - x(a + bi) + R(a + bi) = x2 - Rx - ax - (bi)x + Ra + R(bi)

By combining like terms we get:

x2 - (R + a + bi)x + R(a + bi)

This equation does not have real coefficients which contradicts our original assumption. Thus, no such quadratic equation exists.

Extending to Polynomial Equations

Known Theorems

Using the fundamental theorem of algebra in conjunction with the Complex Conjugate Root Theorem, we see that any odd degree polynomial must have an odd number of real roots (1, 3, 5, ...) and that an even degree polynomial must have an even number of real roots (0, 2, 4, ...).

Proof by Contradiction

Let's take x = -(a+bi) to be a solution where a and b are both real numbers (b is non-zero) and i is the square root of -1. Now represent the real solution with x = -R (R is a real number).

Now the polynomial out in a factored form:

((x+(a+bi))^n)((x+R)^m) where n > 0 and m > 0.

Now multiply out all of the non-real factors collectively and all the real numbers collectively, but leave them separated.

[xn + ... + ((n choose 1)(a+bi)n-1)x + (a+bi)n][xm + ... + Rm]

Case 1: R is non-zero

Now multiply all the terms together:

xn+m + ... + (Rm-1(a+bi)n + Rm(n choose 1)(a+bi)n-1)x + Rm(a+bi)n

If n is an even integer greater than 0 and a = 0, then (bi)n is a real number and (bi)n-1 is an imaginary number.

If n is an odd integer greater than 0 and a = 0, then (bi)n-1 is a real number and (bi)n is an imaginary number.

If a is non-zero, then we have (a+bi)n = (an + ... + (n choose 1)(bi)n-1(a) + (bi)n). Rm(a+bi)n is non-real.

If R = 0, then we have:

xn+m + ... + ((n choose 1)(a+bi)n-1)xm+1 + (a+bi)nxm

and (a + bi)n will always have a non-real part.

Thus, in all cases, the polynomial does not have real coefficients which contradicts our original statement. So no such polynomial exists.