# Irrational and Complex Roots

Reference > Mathematics > Algebra > PolynomialsIf 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 x^{2} + 2x - 1 = 0, which you can solve with the quadratic formula. You obtain the following roots:

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

**Solution:**3 -

**Example:**One root of a rational polynomial equation is

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

**Example:**If x

^{3}- 3x

^{2}- 12x + k = 0, with k rational, and one of the roots is -1 +

**Solution:**If -1 +

^{2}, divided by the leading coefficient, so the sum of the roots is 3. Thus -1 +

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 x^{2} + 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 x^{2} - 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 x^{3} - 7x^{2} + 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: 2x^{3} - 14x^{2} + 34x - 30.

## Questions

^{3}- 8x - k = 0 is 1 -