I would just set your new formula equal to the quadratic formula and simplify.  If you end up with an identity, it is true.  If you end up with something that is obviously not true, then your new formula is not equivalent to the quadratic formula.

x= ((-b+-sqrt{b^2-4ac})/2a

This should become:

x = -b  +/- {b^2 - 4ac} / 2a ........eq. 101

wheras the actual quadratic solution is:

x= (-b/2a) +/- {b^2 - 4ac} / 2a

Therefore:

if:  2a= 1 .........then eq. 101 can be TRUE!

to disprove it since your equation should represent roots of some function f(x) check and see if it works for say
say 3 or 4 functions if you see it seems to work then you can work it out in more algebraic terms

"IF you have x = 2c / [-b +- (b^2 - 4ac)]" ... so what? Simply coincidence, right?

Works in infinite cases, like:
x^2 + 2x + 1 = 0
x^2 + 6x + 9 = 0
4x^2 + 6x + 2 = 0
46x^2 + 83x + 37 = 0

Again: so what?

Thanks a lot for your answers!

So is this solving equation always right for solving? x=  Or is this one wrong? I tried to set them equal to each other but I wasn't able to solve it very well.

Sorry for my bad language but I'm an exchange student for this school year.

I need to prove that it doesn't work. But I don't know how.

Thanks a lot

have  a great day

Marcus

Do what Loren said...

Lets let d=-b+/-sqrt(b^2-4ac)

so if they are the same then the following must be true
d/(2a)=2c/d
d^2=4ac       
d=2sqrt(ac)   we'll assume that ac>0...if the formula always works it should work if this assumption is true

-b+/-sqrt(b^2-4ac)=2sqrt(ac)
+/-sqrt(b^2-4ac)=2sqrt(ac)+b
square both sides
b^2-4ac=4ac+4bsqrt(ac)+b^2
0=8ac+4bsqrt(ac)
0=2ac+bsqrt(ac)
-2ac=bsqrt(ac)
4a^2c^2=b^2(ac)
4ac=b^2
b^2-4ac=0

So the formula will work if b^2-4ac=0 and ac>0


so to disprove this pick a case where this doesnt happen...a=1 b=-3 c=1
x^2-3x+1=0
quad formula gives
(3+/-sqrt5)/2
your formula gives
2/(3+/-sqrt5)
and these are not the same

Your formula is not correct.  There should still be a square root in the denominator.

Try it out on 4*x² + 3*x = 0.  What you have is a barely altered Quadratic Formula, sort of with a "rationalized" numerator.  It is not magic, new, impressive, or in possession of mystic qualities.

For the general quadratic, a*x² + b*x + c = 0, the "old" Quadratic formula has a deficiency for a = 0.  That is why the general quadratic normally is stated as above but with a ≠ 0.  Then the "old" formula has no deficiency at all.  This "new" formula has significant deficiencies for a=0 or c=0.

Nice try, though.  Where did you get it?

Ahhh, I see...the divisor is not -b +- (b^2 - 4ac) but -b +- sqrt(b^2 - 4ac); you gave me a headache, Marcus !

bri says:
"so to disprove this pick a case where this doesnt happen...a=1 b=-3 c=1
x^2-3x+1=0
quad formula gives
(3+/-sqrt5)/2
your formula gives
2/(3+/-sqrt5)
and these are not the same"

Disagree; only a "+-" rearrangement :
(3 + sqrt(5)) / 2 = 2 / (3 - sqrt(5))
(3 - sqrt(5)) / 2 = 2 / (3 + sqrt(5))

Only problem I can see is with c=0 (a=0 has same problem in both formulas).

So I now prefer this new formula!
x = 2c / (d - b) where d = +-sqrt(b^2 - 4ac) and ac <> 0

x= ((-b+-sqrt{b^2-4ac})/2a


The range of possibilities seem to be:


a=.5

-< B <

- < C <  

While B & C can't both equal zero, either.

Are they equal?

[-b + sqrt(b^2-4ac)] / 2a = 2c / [-b - sqrt(b^2-4ac)]

4ac = [-b + sqrt(b^2-4ac)][-b - sqrt(b^2-4ac)]

4ac = b^2 - (b^2 - 4ac)

4ac = b^2 - b^2 + 4ac

4ac = 4ac

Same thing with:
[-b - sqrt(b^2-4ac)] / 2a = 2c / [-b + sqrt(b^2-4ac)]

They're equal !

...ignoring Domain issues.

Do ma in ? That's matricide :)

Thanks a lot for your answers!

So is the other formular just written in a different way? Or does the other one has the two restrictions: a cannot equal 0 and c cannot equal 0?

Thanks a lot!

Ps. what does matricide means?

Greetins Marcus

The "other" formula is a logical result of "completing the square" with the general quadratic.  It is "the" quadratic formula.

Your "new" version is from an illogical additional step in the process, deliberately adding additional domain restrictions.  Why would one want to do that?  Throw it away.

"While B & C can't both equal zero, either."

If C ≠ 0, it does not define additional information to state what you have stated.

Solve: x^2 + 4x + 3 = 0

No.

  Misunderstood question.

You People are Great with Quadratic Formulas !!!
But what about Angle Bisectors ???
Is a Triangle uniquely determined by its three Angle Bisectors ??
I proved it can for the Isosceles !
Hello  Rocky ! Where are you this year !