I'm given the following quadrating equation:
$ax^2+bx+c = 0$
and im asked to find a c such that there is only one solution to the equation.

I know that the solutions to a quadratic equation can be found with this formula:
$x = \frac{(-b \pm \sqrt{ b^2 - 4ac } )}{2}$
which can have 2 solutions $x_-$ and $x_+$

so for there to be just one solution $x_-$ must equal $x_+$
so i took this and derived the following:
$<br /> 1) x_- = x_+<br />$

$<br /> 2) \frac{(-b + \sqrt{ b^2 - 4ac } )}{2} = \frac{(-b - \sqrt{ b^2 - 4ac } )}{2}<br />$

$<br /> 3) -b + \sqrt{ b^2 - 4ac } = -b - \sqrt{ b^2 - 4ac }<br />$

$<br /> 4) \sqrt{ b^2 - 4ac } = - \sqrt{ b^2 - 4ac }<br />$

$<br /> 5) b^2 -4ac = -b^2 + 4ac<br />$

$<br /> 7) 2b^2=8ac<br />$

$<br /> 8) c = \frac{2b^2} {8a}<br />$

$<br /> 9) c = \frac{b^2} {4a}<br />$

and that actually works. For any quadratic equation with the given values a and b i can find a c such that the equation only has one solution. The problem is, in retrospect i have no idea why this is true because i completely fucked up that derivation. Mainly in step 5 where i raised both sides to the power of 2 to remove the square roots, but i forgot that, that should also makes the "-" disappear from the right side and so it should just equal itself and be unsolvable from there. However because i forgot about it i ended up deriving something that is still true? how can that be?

also before that in step 4 i have sqrt(something) = - sqrt(something), how can that be true? as far as i can tell i didn't do anything wrong algebraically up until then, and the initial equation should be true right?