|T O P I C R E V I E W
||Posted - 09/25/2008 : 08:52:23
The following information was given for the following problem:
mx + nx + 16 = 0
In the quadratic equation above, m and n are nonzero numbers. If the equation has only one solution, which of the following is equal to m in terms of n?
The following answers given were:
(a) m = n/2
(b) m = n/4
(c) m = n/16
(d) m = n/16
(e) m = n/ 64
I need an explanation on how to solve this type of problem. Any assistance would be greatly appreciated.
|2 L A T E S T R E P L I E S (Newest First)
||Posted - 09/25/2008 : 14:35:01
Thanks again Skeeter. I thought about the general equation but for some reason the "m' and "n" threw me for a curve. Thanks again for all your help.
||Posted - 09/25/2008 : 11:17:53
the general quadratic is ...
y = ax+bx+c
if y = 0, then x = [-b [sqrt(b-4ac)]/(2a)
further, if the quadratic equation has a single solution, then the discriminant, b-4ac, equals 0.
in your problem ...
a = m
b = n
c = 16
work with that info.