Question

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

Answer 1

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 `bne0` as well as when `b=0` to prove it.

Explanation:

Let's write out the quadratic formula first:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

This assumes the equation looks like: `ax^2+bx+c` in this scenario:

`a=1`
`b=a`
`c=b`

Filling out the equation:

`x=(-a+-sqrt(a^2-4b))/(2)`

Now, if `bne0`, the solution for the zeroes can be given using the above equation. However, if `b=0`, we can move forward:

`x=(-a+-sqrt(a^2-4(0)))/(2)=(-a+-sqrt(a^2))/(2)`

`x=(-a+-a)/2`

`x={0,-a}`