# How do you find the complex roots of x^2-14x+49=0?

Jan 23, 2017

See below.

#### Explanation:

Assuming complex their roots, the polynomial can be represented as

$\left(x + a + i b\right) \left(x + a - i b\right) = {\left(x + a\right)}^{2} + {b}^{2} = {x}^{2} + 2 a x + {a}^{2} + {b}^{2} = 0$

Comparing

$\left\{\begin{matrix}2 a = - 14 \\ {a}^{2} + {b}^{2} = 49\end{matrix}\right.$

so

$a = - 7$ and $b = 0$

so the polynomial is ${\left(x + 7\right)}^{2} = 0$ with real roots.