Linear and quadratic (Posted on 2007-10-23)

	Submitted by Kurious
Rating: 3.0000 (3 votes)
Solution:	(Hide)
We notice b must be non-zero so that there is an equation to solve. x = -b +/- sqrt(b^2 - 4ac) /2a As we substitute a=0; the denominator becomes 0, and the numerator becomes 0 on +sqrt and diverges on -sqrt. Since the +sqrt version yields 0/0, we instead use: x = lim a->0 -b + sqrt(b^2 - 4ac) /2a which trigges the L'Hospital rule. Because the limit depends on a, we differentiate the numerator and denominator by a. This yields x = lim a -> 0 1/2 * (b^2 - 4ac)^(-1/2) * (-4c) /2 x = 1/2 * b^-1 * -2c x = -c/b as expected.

	Subject	Author	Date
	Two Non Calculus Methods	K Sengupta	2007-11-04 00:08:00
	My 3 solutions.	Jer	2007-10-24 08:31:53
	re: By limits	Brian Smith	2007-10-24 01:59:49
	By limits	Old Original Oskar!	2007-10-24 00:08:37
	re: No Subject	Daniel	2007-10-23 18:12:55
	No Subject	Charlie	2007-10-23 14:54:06
	re(2): Easy Trick Solution	Mike C	2007-10-23 14:07:30
	re: Easy Trick Solution	Matt	2007-10-23 13:36:54
	Easy Trick Solution	Brian Smith	2007-10-23 10:46:57