quadratic formula

(thing) by antizeus Sat Nov 13 1999 at 9:44:51
The quadratic formula is used to find roots of second-degree ("quadratic") polynomials, i.e. expressions of the form a * x^2 + b * x + c .

For the above polynomials, the roots are: x = (-b +/- sqrt(b^2 - 4ac)) / (2a)

(thing) by blaaf Sat Sep 23 2000 at 7:24:08
Proof (completing the square method):

ax2 +bx + c = 0
ax2 + bx = -c
x2 + bx / a = -c/a
x2 + bx / a + (b / 2a)2 = -c / a + (b / 2a)2
(x + (b / 2a))2 = -4ac / 4a2 + b2 / 4a2
x + (b / 2a)2 = (b2 - 4ac) / 4a2
x + (b / 2a) = +/- sqrt ((b2 - 4ac) / 4a2)
x = -b / 2a +/- sqrt(b2 - 4ac) / 2a
x = -b +/- sqrt(b2 - 4ac) / 2a

(thing) by Shoegazer Sat Sep 23 2000 at 18:04:51
In my Algebra 2 class, our crazy teacher taught us a song where you sing the quadratic formula to the tune of "Pop Goes The Weasel".

X equals the opposite of B
Plus or minus the square root
of b squared minus 4 A C
All over 2A!

I can't tell you why it worked so well, but to this day, I can never forget it...
(idea) by jafuser Sat Sep 23 2000 at 18:38:02
Most commonly written as something somewhat resembling:
           .----------
 -b +  _  /  b2 - 4ac
        \/             
----------------------
          2a
My stupidity now corrected thanks to JustSomeGuy
(idea) by bob the cow Tue Oct 24 2000 at 0:09:23
If ax2 + bx + c = 0, then:
        -b + root( b2 - 4ac )
 x = ----------------------------
            2 a

Want to use this on your fancy TI-83? Use this program:
:a+bi
:Lbl 1
:ClrHome
:Input "A=",A
:Input "B=",B
:Input "C=",C
:If A=0:Goto 1
:(-B+root(B2-4AC))/(2A) -> X
:(-B-root(B2-4AC))/(2A) -> Y
:Disp "ANSWERS STORED"
:Disp "AS X AND Y"
:Disp X
:Disp Y
Where -> indicates the "store" operator, a+bi indicates the command that sets a+bi mode and root() indicates the square root operator.
(thing) by General Wesc Fri Mar 16 2001 at 21:07:15
The proof of the quadratic equation is delightfully simple to learn. It's also the only proof I had to memorise for College Algebra:


                   Start with: ax2 + bx + c = 0

                   Subtract c: ax2 + bx = -c

                  Divide by a: x2 + bx/a = -c/a

          Complete the Square: x2 + bx/a + b2/4a2 = -c/a + b2/4a2

   Factor left, combine right: (x+b/2a)2 = (b2-4ac)/4a2

                  SQRT it all: x+b/2a = ±SQRT(b2-4ac)/2a

Subtract b/2a from both sides: x = [-b ±SQRT(b2-4ac)]/2a

Sometimes called the quadratic formula. (Probably a better name since it's used to solve quadratic equations.)

(idea) by mattbw Sun Aug 17 2003 at 21:10:09
The numerical properties of the Quadratic Equation make it unsuitable for general use in software. For example, consider the special case a=0, which a human would obviously recognize. For a computer, special code must be written. More subtly, from Numerical Analysis, we must consider the possibility of cancellation error when b2 is much greater than 4ac. In this case, the solution for one root will be very close to zero, and the other will be close to -b/a. The near-zero number will suffer accuracy problems.

To avoid the difficulty with cancellation, we may compute the roots with the following:

z = -0.5 (b + sign(b)*(b2 - 4ac))

x1 = z/a
x2 = c/z

Source: Numerical Recipes
(idea) by m8888888 Sat Apr 24 2004 at 21:48:42

In my Algebra 2 class, we learned the quadratic formula to the tune of "Gillligan's Island":
Negative B plus or minus
The square root of
B-squared minus 4ac
All over 2a
All over 2a

Other people claimed to have learned the handy formula by singing it to the tune of:
Frère Jacques
Pop Goes the Weasel
Row, Row, Row Your Boat
God Rest Ye Merry Gentlemen
Notre Dame Fight Song
U of M Fight Song
Jingle Bells
Ballin' the Jack
Amazing Grace

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