They probably derived them from existing rules.

by Anonymous 11 years ago

http://ctrlv.in/94028 ax^2 + bx + c = 0 ax^2 + bx = -c a(x^2 + bx/a) = -c a(x^2 + bx/a + b^2/4a^2) - b^2/4a^2 = -c a(x + b/2a)^2 - b^2/4a^2 = -c a(x + b/2a)^2 = b^2/4a^2 - c a(x + b/2a)^2 = (b^2 - 4ac)/4a (x + b/2a)^2 = (b^2 - 4ac)/4a^2 x + b/2a = +-(√b^2 - 4ac)/2a x = -b/2a +- (√b^2 - 4ac)/2a x = (-b +- √b^2 - 4ac)/2a This took me about 5 minutes.

by Anonymous 11 years ago

Fair enough; nice proof. Now see if you can invent me a new formula in 'about 5 minutes' that works in the vast majority of cases.

by Anonymous 11 years ago

It's not the vast majority; it has to work in all cases to be accepted.

by Anonymous 11 years ago

There can be exceptions - with the quadratic formula, it won't work if a is equal to zero.

by Anonymous 11 years ago

We don't need a new formula; this one works fine. If you're referring to a formula for something different, then you can't just say "Give me a new formula!!!" you need to tell me "Give me the formula for..." Also, I'm not a mathematician, so I'm not the go-to person for new formulas. a = 0 isn't really an exception. If a = 0, then ax^2 = 0, and what remains (bx + c = 0) is not a quadratic equation. Edit: I actually made a mistake when typing that proof (I solved it on paper a while ago, I didn't just work it out now). Equations 4-6 should say b^2/4a (not b^2/4a^2) because it was multiplied by 'a' to balance the equation.

by Anonymous 11 years ago

If A = 0 then it's not a quadratic function...

by Anonymous 11 years ago

Most mathematical formulae are developed in the basis of theory and theory only. It takes years of mind numbing calculations but there is no testing involved till the formula is completed.

by Anonymous 11 years ago

It's just completing the square with variables instead of numbers...

by Anonymous 11 years ago