I am using Excel 2000
It is possible to use the quadratic formula for quadratic equations of the
form ax^2 bx c. Solving for values of x use: (-b or -(sqroot(b^2-4ac))/2a
quot;Cameron PEquot; wrote:
gt; I am using Excel 2000
Nice in theory, bad idea in practice. Quoting (with slight editing for
this context) from Press, et al, /Numerical Recipes in C/:
lt;quotegt; If either a or c (or both) are small, then one of the roots will
involve the subtraction of b from a very nearly equal quantity (the
discriminant); you will get that root very inaccurately. The correct way
to compute the roots is
q = -(b sgn(b)*sqrt(b^2-4a*c))*0.5
Then the two roots are
x1 = q/a and x2 = c/q
lt;end quotegt;
You can get this straight off of the web by searching for quot;Numerical
Recipes in Cquot;, then look for Chapter 5. section 6 (Quadratic and Cubic
Equations), pp 183 ff.
Harlan Grove and I (and perhaps Jerry Lewis) had a thread on this some
time back. I wrote the equations above to accommodate VBA to an extent.
You may be able to rearrange things to make for a snappier function
call, but you get the idea (I hope).
Regards,
Dave Braden
CJ wrote:
gt; It is possible to use the quadratic formula for quadratic equations of the
gt; form ax^2 bx c. Solving for values of x use: (-b or -(sqroot(b^2-4ac))/2a
gt;
gt; quot;Cameron PEquot; wrote:
gt;
gt;gt; I am using Excel 2000
--
Please keep response(s) solely within this thread.
- Feb 22 Thu 2007 20:35
how do I use the solver to solve polynomial equations?
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