# How do you show that one of the roots of x^2+ax+b=0 is a, if and only if b=0?

Mar 27, 2018

First of all, the root would be $x = - a$. Treat the equation as if it is a full quadratic, and then attempt to solve when $b \ne 0$ as well as when $b = 0$ to prove it.

#### Explanation:

Let's write out the quadratic formula first:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

This assumes the equation looks like: $a {x}^{2} + b x + c$ in this scenario:

$a = 1$
$b = a$
$c = b$

Filling out the equation:

$x = \frac{- a \pm \sqrt{{a}^{2} - 4 b}}{2}$

Now, if $b \ne 0$, the solution for the zeroes can be given using the above equation. However, if $b = 0$, we can move forward:

$x = \frac{- a \pm \sqrt{{a}^{2} - 4 \left(0\right)}}{2} = \frac{- a \pm \sqrt{{a}^{2}}}{2}$

$x = \frac{- a \pm a}{2}$

$x = \left\{0 , - a\right\}$