Irrational and Complex Roots

Reference > Mathematics > Algebra > Polynomials

If a polynomial equation has all rational coefficients, then we know something important about that equation's irrational roots. They come in pairs. Consider the quadratic equation x² + 2x - 1 = 0, which you can solve with the quadratic formula. You obtain the following roots:

x = 1 +

or x = 1 -

Do you notice anything interesting about those two roots? They have the same rational part, and their irrational parts are opposites of each other. That makes sense, doesn't it? Considering what you know of the quadratic equation, it seems reasonable that this would always happen. And it does. Not only that, but it happens for higher degree polynomials as well. If a +

is a root, then so is a -

Example: Two of the roots of a rational polynomial equation are 3 +

and 2 -

. What are two other roots?

Solution: 3 -

and 2 +

Example: One root of a rational polynomial equation is

. Can you conclude anything about other roots of the equation?

Solution: You can think of the given root as 0 +

. Thus, another root must be 0 -

, or just -

Example: If x³ - 3x² - 12x + k = 0, with k rational, and one of the roots is -1 +

, find the value of k.

Solution: If -1 +

is a root, then -1 -

is also a root. Let m be the third root. We know that the sum of the roots must be the opposite of the coefficient of x², divided by the leading coefficient, so the sum of the roots is 3. Thus -1 +

- 1 -

+ m = 3. Conveniently, the square roots cancel, leaving -2 + m = 3, or m = 5.

k = -(5)(-1 +

)(-1 -

) = -5(1 - 3) = -5(-2) = 10.

If a polynomial equation has real roots, then any non-real solutions also come in pairs, according to the same pattern; the two complex roots will be conjugates of one another.

Consider the equation x² + 2x + 2 = 0. If you solve this with the quadratic formula, you will find that the roots are:

x = 1 + i and x = 1 - i. The real parts match, and the imaginary parts are opposite.

Example: A quadratic equation with no leading coefficient has 1 + 2i as a root. Find the equation.

Solution: The other root is 1 - 2i. The sum of the roots is therefore 1 + 2i + 1 - 2i = 2, and therefore the coefficient of x is the opposite of that, or -2. The product of the roots is (1 + 2i)(1 - 2i) = 1 + 4 = 5. This is the constant term. Thus, the equation is x² - 2x + 5 = 0

Example: Two of the zeroes of a cubic polynomial are 3 and 2 - i, and the leading coefficient is 2. What is the polynomial?

Solution: The other root is 2 + i. We can simply multiply together the factors (x - 2 - i)(x - 2 + i)(x - 3) to obtain x³ - 7x² + 17x - 15. Since this does not have a leading coefficient of 2, we can multiply the entire polynomial by 2 to obtain the final result: 2x³ - 14x² + 34x - 30.