# finding x in quadratic equations

• Dec 24th 2009, 07:55 AM
renlok
i know the equation of $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find if at all where a function crosses the x axis but i also found you can just use $\displaystyle x = \frac{-b}{2a}$ which i really dont understand as were does the $\displaystyle \sqrt{b^2 - 4ac}$ go?
• Dec 24th 2009, 08:03 AM
galactus
$\displaystyle x=\frac{-b}{2a}$ gives the x coordinate of the vertex of a parabola.
• Dec 24th 2009, 08:03 AM
skeeter
Quote:

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

$\displaystyle x = \frac{-b}{2a}$ 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 $\displaystyle \pm \frac{\sqrt{b^2-4ac}}{2a}$ on either side of $\displaystyle x = \frac{-b}{2a}$
• Dec 24th 2009, 08:04 AM
e^(i*pi)
Quote:

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

That is only true when $\displaystyle b^2-4ac$ (which is called the discriminant) is equal to $\displaystyle 0$

edit: see above
• Dec 24th 2009, 08:12 AM
renlok
oh right that makes sense thanks guys :D