I didn't post that; you around, Dr See ?

Subject:

Comments:
: : 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

Optional Link URL:
Link Title:
Optional Image URL: