I've come up with my own quadratic equation

[DavidNyan10]

I know I'm 8 years too late [for this thread], but I've came up with my own quadratic equation:

$$ax^2+bx+c\:=\:0\hspace{1in}a,\:b,\:c\:\in\:\mathbb{R} \\\\ x\:=\:-\frac{c}{b}+a\left[\frac{-b\pm\sqrt{b^2-4ac}}{2a^2}+\frac{c}{ab}\right]$$

 

[stapel]

I've came up with my own quadratic equation:

$$ax^2+bx+c\:=\:0\hspace{1in}a,\:b,\:c\:\in\:\mathbb{R} \\\\ x\:=\:-\frac{c}{b}+a\left[\frac{-b\pm\sqrt{b^2-4ac}}{2a^2}+\frac{c}{ab}\right]$$

Do you perhaps mean to say that you have invented your own Quadratic Formula? If so, how do you feel that what you have posted is new or distinct? (Note: It simplifies to the "regular" Quadratic Formula, so I'm not seeing how your "equation" is different from the regular Formula.)

 

[topsquark]

I know I'm 8 years too late [for this thread], but I've came up with my own quadratic equation:

$$ax^2+bx+c\:=\:0\hspace{1in}a,\:b,\:c\:\in\:\mathbb{R} \\\\ x\:=\:-\frac{c}{b}+a\left[\frac{-b\pm\sqrt{b^2-4ac}}{2a^2}+\frac{c}{ab}\right]$$

This is not correct.

-Dan

 

[Steven G]

I know I'm 8 years too late [for this thread], but I've came up with my own quadratic equation:

$$ax^2+bx+c\:=\:0\hspace{1in}a,\:b,\:c\:\in\:\mathbb{R} \\\\ x\:=\:-\frac{c}{b}+a\left[\frac{-b\pm\sqrt{b^2-4ac}}{2a^2}+\frac{c}{ab}\right]$$

Can you please explain how you can use your formula when b=0. I seem to be confused.

For the record, you can take any formula, add and subtract the same value and/or multiply and divide by the same number and then claim that you now have a different formula.. The question is whether the formula is different from the original one.

May I ask what the point is that you are trying to make? Also, why are you 8 years too old for this post??

 

[stapel]

...why are you 8 years too old for this post??

This question was original appended to an 8-year-old thread. Apparently the link to that old thread did not survive the splitting. (??)

Apologies for the confusion. ?

 

[topsquark]

This question was original appended to an 8-year-old thread. Apparently the link to that old thread did not survive the splitting. (??)

Apologies for the confusion. ?

@stapel
@DavidNyan10

Does the original thread explain why a "new" form of the quadratic formula would be useful?

-Dan

 

[stapel]

@stapel
Does the original thread explain why a "new" form of the quadratic formula would be useful?

If memory serves, the original thread was about using the Quadratic Formula.