The plus-or-minus threw me for a loop at first (so I digressed with other methods), but of course you can just rationalize for each root individually and they switch which one is plus and which one is minus.
This isn’t nearly as useful as the more commonly-known formula. For the commonly-known one, the only requirement is that a≠0, which isn’t a big deal because if a=0 then the equation isn’t meaningfully a quadratic. But for this given one, if c=0 it breaks down. Consider x²–x=0; this clearly can be factored as x(x-1)=0 and so has solutions x=0 and x=1. Plugging in a=1, b=-1, c=0 to the given formula results in x=0/(1±1); i.e., x=0/2=0 and x=0/0. That’s no good.
You can work out how to use it usefully in the sine of a sum/difference as well. I learned it long, long ago (1980?) from my 11th grade Analysis teacher (that’s what we called precalc w/trig course back then).
Very interesting. Never thought quadratics has alternative formula for the roots.
Have you ever seen the quadratic formula proved in the following way???
After moving c to the other side, multiply both sides by 4a. Then complete the square, but not in the usual way. Intuitively add b^2 to both sides. The rest is easy. It avoids fractions in the derivation.
Actually both formulas work even when the divisor is zero.
Example ax^2+bx+c= 0 when a = 0
you take derivative with respect to a for case of 0/0 of both parts
that is when (b – sqrt(b^2-4ac))/2a when a =0 rewriting as
it becomes (b – (b^2-4ac)^(.5))/(2a) then taking derivatives with respect to a you get
(-.5(b^2 -4ac)^(-,5)(-4c))/2 at a = 0 you get -c/b you actually also
get c/b since sqrt has two solutions so when you plug back in
only one works. The other case when you take + sqrt at start
give a nonzero number divided by zero which is a singular and
in algebra is not considered a number. So again with a = 0 the standard formula gives the correct answer even if a = 0
In the alternate formula for solutions when a = 0 there is again the
same problem when a=0 but the aquation is not in indeterminate form
so no derivatives needed in its evaluation. One leads directly to
-c/b the other to a the singularity which is not counted as a solution.
So at least for a = 0 or near zero the second solution is usually the better one to implement. However the thing looks nasty at c= 0 and
have for the divisor b-sqrt(b^2+4ac) at the plus sqrt you see x = 0
as one of the solutions. for the current case do as in first example
take derivatives set c=0 and you can get the other root no problem.
The alternate form is better written as:
-2c/(b +-sqr(b^2 – 4ac))
Where (-b +-sqr(b^2 – 4ac))/(2a) = -2c/(b +-sqr(b^2 – 4ac))
So when the b value is +
x1 = -2c/(b + sqr(b^2 – 4ac)) this form is more accurate here
x2 = (-b – sqr(b^2 – 4ac))/(2a)
And when the b value is –
x1 = (-b + sqr(b^2 – 4ac))/(2a)
x2 = -2c/(b – sqr(b^2 – 4ac)) this form is more accurate here