Where does the quadratic formula come from

I've never had a teacher who's explained the rationale behind the quadratic formula. I tried to work backwards to figure out (it's kind of hard to read in this format):

x=(-b+/-sqrt(b^2-4ac))/2a
2ax=-b+/-sqrt(b^2-4ac)
2ax+b=+/-sqrt(b^2-4ac)
4a^2x^2+4abx+b^2=b^2-4ac
4a^2x^2+4abx=-4ac
4a^2x^2+4abx+4ac=0
4a(ax^2+bx+c)=0

That's where I'm stuck. The general form for a quadratic equation is there, but what's with the 4a? Am I even on the right track?

The 4a is much easier to see from the other way. Start with:
ax^2 + bx + c = 0

Bring c to the other side:
ax^2 + bx = -c

Divide by a on all sides:
x^2 + b/a * x = -c/a

Complete the Square:
x^2 + b/a * x + b^2/4a^2 = -c/a + b^2/4a^2

Factor the Left Hand Side, and Add the terms on the right hand side:
(x+b/2a)^2 = -4ac/4a^2 + b^2/4a^2
(x+b/2a)^2 = (b^2-4ac)/4a^2

Square root both sides:
x + b/2a = ±√(b^2-4ac) / 2a

Bring everything to one side and isolate the x:
x = (-b ±√ (b^2 - 4ac))/2a

Hi:

you are doing is: completing the square with Algebraic Variables instead of numbers

expect that you are using the end result as a short cut instead of doing all the steps involved

and the 4ac is called a discriminant which tell you what type of roots you will get depending of what value the discriminant is as follows:


If the discriminant is positive, then there are two distinct roots, both of which are real numbers

If the discriminant is zero, then there is exactly one distinct real root, sometimes called a double root:

If the discriminant is negative, then there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other


for more info

http://en.wikipedia.org/wiki/Quadratic_e…

http://en.wikipedia.org/wiki/Quadratic_e…

http://mste.illinois.edu/exner/ncsa/quad…

http://mathworld.wolfram.com/QuadraticFo…

http://www.purplemath.com/modules/quadfo…

I hope this helps

ax² + bx + c = 0

First divide through by a then complete the square and solve for x

x² + b/a x + c/a = 0

x² + b/a x + (b/2a)² = –c/a + (b/2a)²
(x + b/2a)² = –c/a + (b/2a)²
x + b/2a = ±√(–c/a + b²/4a²) = ±√[(b² – 4ac)/4a²] = ±√(b² – 4ac)/2a

x = [–b ± √(b² – 4ac)]/2a

since anything *0 is still 0 the ax^2+bx+c=0 makes that true even with the 4a. if a is 0 then the formula doesn't work because you can't divide by 0.