finding x in quadratic equations

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December 24th 2009, 07:55 AM
renlok

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?
December 24th 2009, 08:03 AM
galactus

$x=\frac{-b}{2a}$ gives the x coordinate of the vertex of a parabola.
December 24th 2009, 08:03 AM
skeeter

Quote:

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 $x = \frac{-b}{2a}$
December 24th 2009, 08:04 AM
e^(i*pi)

Quote:

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
December 24th 2009, 08:12 AM
renlok

oh right that makes sense thanks guys :D