How did they get the quadratic equation?

Question

How do you get from ax^2+bx+c=0 to x= -b + or - sqrt (b^2-4ac)/2a?

Answer 1

ax^2 + bx+c = 0
ax^2 + bx = -c
x^2+(b/a)x+ ... = -c/a + ...
x^2+(b/a)x+ (b/2a)^2 = -c/a + (b/2a)^2
{x - (b/2a)}^2 = -4ac/4(a^2) + b^2/4(a^2)
x - (b/2a)}^2 = + or - root{(-4ac + b^2)/4(a^2)}
x= {-b + or - sqrt (b^2-4ac)}/2a

Th

Answer 2

You can solve any quadratic equation by a method called "completing the square." This includes the GENERAL quadratic equation, ax^2+bx+c=0. If you apply the method of completing the square to ax^2+bx+c=0, you get the "quadratic formula" (which you accidentally called the "quadratic equation" in your question).

Here are the steps of the process:

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

Answer 3

I believe it was derived by the method of "completing the squares". Look that up. I remember having to drive the quadratic equation in high school

Answer 4

Boy that question would be SO much easier to answer if we could actually use mathematical symbols...

Suffice it to say that one equation undoes the other, much like, on a simpler scale, subtraction undoes addition or division undoes multiplication.

Answer 5

Start with the general quadratic you have above and complete the square.

http://mathworld.wolfram.com/QuadraticEquation.html