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?

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!

On Apr 29, 2:49 pm, Paul <pkshree...@gmail.com> wrote:
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

I'm puzzled. Shouldn't these be
X = -b -(-4ac+b^2)^(1/2)/(2a) , -b +(-4ac+b^2)^(1/2)/(2a)

Everyone,

Thank you very much! Please accept my apologies for such silly
mistakes of writing wrong solutions to usual quadratic and even linear
equations! I was looking at too many things at once while typing that
question, (some specialized equations, some general equations's
solutions, etc) so I ended up mixing them up in this post.

Thank you for correcting me! The book by Newcomb and reply by Dave
gave me very interesting alternative solution to quadratic equation
(one that I've never seen) that allowed me to do the limit function
toward linear limit. However I find it interesting that it still have
"sudden death" at the zero (two solutions suddenly to one solution).
While I will continue to work on this transistion issues, I want to
"promote" this discussion, the problem is the following,

partial differiental equations:

{dX}/{dT} = a X^2 + b X + c :: X and T is separable, and variable
"a" that goes from 1 to the limit of zero (as earlier question,
however now with dX/dT).
Now, we can solve this simply by integrating each side separably,

{dX}/{a X^2 + b X + c}= dT = Delta T

(oh hell, I'll just paste entire equation here in LaTeX typeset, I
don't know if it is possible to directly paste or attach to this
Usenet group?) I integrated it with X_{bottom} and X_{top} to create
"closed recursive" equation.
X_t=\frac{\sqrt{ac-\frac{b^2}{4}}X_b + (c+\frac{b}{2}X_b)\tan
((\sqrt{ac-\frac{b^2}{4}})\Delta T) }{\sqrt{ac-\frac{b^2}{4}}-(a X_b+
\frac{b}{2})\tan ((\sqrt{ac-\frac{b^2}{4}})\Delta T)}

Now, if I allow the variable "a" to go zero, of course this equation
goes nuts. The solution for {dX}/{b X + c}= dT = Delta T is
completely different using same approach as above (exponential
function).

I have tried using this quadratic "closed recursive" equation anyway
and allow the variable "a" to go exceedingly low but it created
absolutely no results. I suspect I have to develop a different method
of solving this, and possibly have to use something like Riemann's
solution or some type of "open" solution that doesn't completely fix
the type of curve for the solution?

Anyone can help me here? (Many thanks for previous remarks/comments/
help!)

Paul

0...however, as you can see, the graph is absolutely non-respondent
to the changing value of "a". Anyone knows why? Here's link to the

On Apr 29, 2:49=A0pm, Paul <pkshree...@gmail.com> 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 =A0 , =A0 -b +(-4a+b^2)^(1/2)/2a

then we have linear equation,
0= b X + c
=A0with 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? =A0the "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? =A0Is this even possible? =A0If 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) ) [1]

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.