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EdDro

  • one year ago

Does anyone kno the quadratic formula??

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  1. rizwan_uet
    • one year ago
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    |dw:1358703755414:dw|

  2. texaschic101
    • one year ago
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    does this help ?.... http://www.purplemath.com/modules/quadform.htm

  3. Yahoo!
    • one year ago
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    \[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

  4. shamim
    • one year ago
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    \[ax ^{2}+bx+c=0\]

  5. shamim
    • one year ago
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    is called quadratic equation

  6. shamim
    • one year ago
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    the solution of this quadratic equation is \[x=\frac{ -b \pm \sqrt{b ^{2}-4ca} }{ 2a }\]

  7. abb0t
    • one year ago
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    \[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] where you have a polynomial in the form \[ax^2 + bx +c = 0\]

  8. AccessDenied
    • one year ago
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    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.

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