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rizwan_uet
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1358703755414:dw

texaschic101
 2 years ago
Best ResponseYou've already chosen the best response.0does this help ?.... http://www.purplemath.com/modules/quadform.htm

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ b \pm \sqrt{b^24ac} }{ 2a }\]

shamim
 2 years ago
Best ResponseYou've already chosen the best response.0is called quadratic equation

shamim
 2 years ago
Best ResponseYou've already chosen the best response.0the solution of this quadratic equation is \[x=\frac{ b \pm \sqrt{b ^{2}4ca} }{ 2a }\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ b \pm \sqrt{b^24ac} }{ 2a }\] where you have a polynomial in the form \[ax^2 + bx +c = 0\]

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.0It is possible to figure out this formula by solving that quadratic equation using the method of completing the squares. It is also interesting to note some small things about the formula that may help remember it easier: \(\displaystyle x = \frac{b}{2a} \pm \frac{\sqrt{b^2 4ac}}{2a} \) The first term is the vertex xvalue. This formula in a sense says that the two roots are a distance of \(\displaystyle \frac{\sqrt{b^2  4ac}}{2a}\) from the vertex, in position on the xaxis. (I mean, the horizontal distance) The discriminant portion \(b^2  4ac\) can also be used as a sort of shortcut to figure out more about the roots you'll find. It is under a square root, so we can tell that only nonnegative values give real solutions. If it is zero, then we see that we are adding + 0 to the vertex; i.e., the vertex is also the one root.
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