Posted by Soroban on September 13, 2002 at 12:50:16:
In Reply to: stupid, stupid quadratics. posted by tony on September 13, 2002 at 04:45:05:
:
: 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?
Tony, you squared wrong. You must "isolate" the radical before squaring.
sqrt(x^2 + 3x) = 6 - x^2 - 3x, and then square. ack!
But there is a simpler way. We are dealing with Quadratics, as you said,
of the form au^2 + bu + c = 0. That is, there is a u and a u^2.
In this problem, there is a sqrt(x^2 + 3x) and a (x^2 + 3x).
Let u = sqrt(x^2+3x). The equation becomes: u^2 + u - 6 = 0
Solve the quadratic: u = 2, y = -3
Then: sqrt(x^2+3x) = 2, and sqrt(x^2+3x) = -3.
And you can solve those two equations (both quadtratic) and produce four answers.
Be sure to CHECK your answers when dealing with "radical" equations.
Quite often, some answers do not check and must be rejected.
: also.. i have one more that i need help with.
: ___1___ = __1__ + 2
: (x+1)2 x+1
Multiply through by (x+1)^2: 1 = (x+1) + 2(x+1)^2
and you have a "normal" quadratic to solve.
: 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)
Here you have x's in the denominator, just like the previous one.
This time, multiply by x^2: 3 - 7x - 6x^2 = 0
Hope that helps.