- #1
mathzeroh
- 98
- 0
Hellooooo everybody. i have a very distinct feeling that there's something very wrong here...
OK, what I'm trying to do is to get to the quadratic equation through the standard form of a polynomial equation in the second degree.
watch this, and go to this website. it lists the process very briefly http://mathworld.wolfram.com/QuadraticEquation.html...in this picture http://mathworld.wolfram.com/qimg558.gif isn't it wrong to put the x+b/2a inside the parenthesis with the WHOLE thing raised to the second power?? i mean doesn't that mean that the WHOLE THING INCLUDING the 'x' gets squared?? that's wrong! aha! that's wrong! i knew it! i knew it! Nah I'm just kidding there but i really do want to know if that's a typo or not. I think that when completing the square, you should only raise the "b" term in (the standard form) to the second power. like this:
[tex]ax^2+bx+c=0[/tex]
[tex](\frac{b}{2})^2[/tex]
subtract the 'c' and bring it over to the other side, then complete the square:
[tex]ax^2+bx+(\frac{b}{2})^2=-c+(\frac{b}{2})^2[/tex]
is this right, as opposed to the ones shown on that website? :uhh: :uhh:
OK, what I'm trying to do is to get to the quadratic equation through the standard form of a polynomial equation in the second degree.
watch this, and go to this website. it lists the process very briefly http://mathworld.wolfram.com/QuadraticEquation.html...in this picture http://mathworld.wolfram.com/qimg558.gif isn't it wrong to put the x+b/2a inside the parenthesis with the WHOLE thing raised to the second power?? i mean doesn't that mean that the WHOLE THING INCLUDING the 'x' gets squared?? that's wrong! aha! that's wrong! i knew it! i knew it! Nah I'm just kidding there but i really do want to know if that's a typo or not. I think that when completing the square, you should only raise the "b" term in (the standard form) to the second power. like this:
[tex]ax^2+bx+c=0[/tex]
[tex](\frac{b}{2})^2[/tex]
subtract the 'c' and bring it over to the other side, then complete the square:
[tex]ax^2+bx+(\frac{b}{2})^2=-c+(\frac{b}{2})^2[/tex]
is this right, as opposed to the ones shown on that website? :uhh: :uhh:
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