Re: Quadratic to Linear solution

On Apr 29, 2:49 pm, Paul <pkshree...@xxxxxxxxx> wrote:
Everyone,

I'd love assistance to this apparently simple problem (perhaps not?)

First, we have this usual quadratic equation,
0=a X^2 + b X + c
which will have solutions of
X = -b -(-4a+b^2)^(1/2)/2a   ,   -b +(-4a+b^2)^(1/2)/2a

then we have linear equation,
0= b X + c
 with single solution
x = 1/2b

What I have to do is develop somehow a "transition" between these two
extreme limits, allowing the value of "a" go from one to zero, and
somehow have these two equations converge?  the "X" is a function, and
I am trying to also make "a" as a function that allows me to explore
the "range" between the limits of quadratic and linear solution
system?  Is this even possible?  If I solve quadratic equation first,
and do limit with a-> 0, I end up with infinity, instead of linear
solution above.

Any help would be greatly appericated!

Correcting the formulas as I noted in an earlier post, you can
transition from one of the solutions of the quadratic to the solution
of the linear equation as follows.

Define sgn(x) = 1 if x >= 0 and -1 if x < 0.

Using sqrt(.) instead of (.)^(1/2), the solution of the quadratic
equation that remains finite as a --> 0 is given by

X = -2c / ( b + sgn(b) * sqrt(b^2 - 4ac) )

whenever b^2 - 4ac >= 0. Further, notice that

lim_a-->0 [-2c / ( b + sgn(b) * sqrt(b^2 - 4ac) )] = -c/b

The other solution of the quadratic equation becomes infinite as a -->
0.

Dave
.

Relevant Pages

  • Re: I have a question about Bakers Percentage
    ... Brian Mailman wrote: ... > It's a quadratic equation. ... I think it may be called a linear equation. ... Prev by Date: ...
    (rec.food.sourdough)
  • Quadratic to Linear solution
    ... First, we have this usual quadratic equation, ... What I have to do is develop somehow a "transition" between these two ... the "range" between the limits of quadratic and linear solution ...
    (sci.math)