Posted by tony on September 13, 2002 at 04:46:14:
okay, i know how to do regular quadratic equations, with the standard ax2 +/- bx +/- c. here's the probmem posed. x2 + 3x + (radical)x2 + 3x(end radical) = 6 here is what i'm doing, please tell me what i'm doing wrong. okay, to get rid of the radical, i square both sides. so it results in x4 + 9x2 + x2 + 3x = 36, and thus, x4 + 10x2 + 3x = 36. so i have three different exponential values for x.. one to the fourth power, one to the second, and one w/o an exponent. what do i do?
also.. i have one more that i need help with.
___1___ = __1__ + 2
(x+1)2 = x+1
oh yeah.. this is the last one. promise.
3x^-2 - 7x^-1 - 6 = 0
(thus, 3x to the -2 power minus 7x to the -1 power, etc)
help!
-tony