# Math Help - finding x in quadratic equations

1. ## finding x in quadratic equations

i know the equation of $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find if at all where a function crosses the x axis but i also found you can just use $x = \frac{-b}{2a}$ which i really dont understand as were does the $\sqrt{b^2 - 4ac}$ go?

2. $x=\frac{-b}{2a}$ gives the x coordinate of the vertex of a parabola.

3. Originally Posted by renlok
i know the equation of $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find if at all where a function crosses the x axis but i also found you can just use $x = \frac{-b}{2a}$ which i really dont understand as were does the $\sqrt{b^2 - 4ac}$ go?
$x = \frac{-b}{2a}$ gives the x-value of the vertex for the graph of the quadratic (a parabola)

the zeros (or roots) of the quadratic function (where it crosses the x-axis) are $\pm \frac{\sqrt{b^2-4ac}}{2a}$ on either side of $x = \frac{-b}{2a}$

4. Originally Posted by renlok
i know the equation of $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find if at all where a function crosses the x axis but i also found you can just use $x = \frac{-b}{2a}$ which i really dont understand as were does the $\sqrt{b^2 - 4ac}$ go?
That is only true when $b^2-4ac$ (which is called the discriminant) is equal to $0$

edit: see above

5. oh right that makes sense thanks guys