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EdDro
Does anyone kno the quadratic formula??
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does this help ?.... http://www.purplemath.com/modules/quadform.htm
\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
is called quadratic equation
the solution of this quadratic equation is \[x=\frac{ -b \pm \sqrt{b ^{2}-4ca} }{ 2a }\]
\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] where you have a polynomial in the form \[ax^2 + bx +c = 0\]
It 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 x-value. 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 x-axis. (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.