Posted by Dennis Borris on September 13, 2002 at 15:32:27:
In Reply to: Re: stupid, stupid quadratics. posted by T.Gracken on September 13, 2002 at 08:55:41:
Here is a stupid quadratic question for you
- what is ax^2 + bx + cx =
: :
: : 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?
: first, use "^" to indicate an exponent. for example, x^2 means "x squared". second, use sqrt( ) for a square root. for example sqrt(2) means square root of 2.
: now, you might try substituting a letter in place of a bunch of symbols.
: For instance, suppose you let A = sqrt(x2+3x)
: then notice that A2 = x2+3x
: now rewrite the problem using A and A2 to get
: A2 + A = 6
: I'll bet you can solve that for A. Then "back substitute". that is, once you get A = a number, replace the A with sqrt(x2+3x) and solve that equation.
: you can also use this technique to solve other equations that have "quadratic forms". for example, try letting A = 1/(x+1) in the next problem you wrote.
: see if that helps.
: : 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