How do you find the complex roots of x+8=0?

Jan 10, 2017

This equation has one real (complex) root, namely $x = - 8$

Explanation:

Given:

$x + 8 = 0$

This is a polynomial equation of degree $1$, also known as a linear equation.

We can subtract $8$ from both sides of the equation to find:

$x = - 8$

This is the only root.

$- 8$ is a Real number, but it is also a Complex number since the Real numbers are a subset of the Complex numbers.

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Footnote

The Fundamental Theorem of Algebra (FTOA) tells us that any non-constant polynomial has a zero in the Complex numbers.

A straightforward corollary of the FTOA, often stated with it, is that a polynomial of degree $n > 0$ has exactly $n$ Complex (possibly Real) zeros.

So a linear equation has exactly one root, a quadratic equation has two, a cubic has three, etc.