(2/ x-3)^2 - 3(2/ x-3) - 4 = 0
2006-09-01 08:12:14 · 6 answers · asked by Siegma H 1 in Science & Mathematics ➔ Mathematics
You see that 2/(x-3) occurs twice in the formula. Let's call it u, then the formula simplies into
u^2 - 3u - 4 = 0
a normal quadratic equation. It can be factored:
(u - 4) (u + 1) = 0
therefore u = 4 or u = -1.
So 2/(x-3) = 4 or 2/(x-3) = -1.
Multiply with x-3:
2 = 4(x-3) or 2 = -1(x-3)
2 = 4x - 12 or 2 = -x + 3
4x = 14 or -x = -1
x = 3 1/2 or x = 1
2006-09-01 08:17:36 · answer #1 · answered by dutch_prof 4 · 1⤊ 0⤋
put 2/(x-3)=t
now the Q.E.is t^2-3t-4=0
=>(t+1)(t-4)=0 or t=-1 or 4
i.e.2/(x-3)=-1 or -x+3=2 or x=1
2/x-3=4 or 4x-12=2
4x=14 or x=3.5
so the solution set is x=1 or 3.5
2006-09-01 08:33:22 · answer #2 · answered by raj 7 · 0⤊ 0⤋
U have to multiply out the brackets 1st
(2/x-3)(2/x-3) - 6/x+9 = 4
i hope this help a bit
2006-09-01 08:21:53 · answer #3 · answered by davethestickman 2 · 0⤊ 0⤋
I think you are missing the purpose of education dude. You're supposed to do your own homework instead of just asking people for the answers. This way you learn to think on your own instead of spending your life relying on others for your intelligence.
2006-09-01 08:19:33 · answer #4 · answered by Sordenhiemer 7 · 0⤊ 0⤋
You need to remove your values from your equation
(2/x-3)^2-3(2/x-3)-4=0
(2/x)^2-(2/x)=(-3)^2 - 3(-3)-4
(2/x)^2-(2/x)=(-9)-10
(2/x)^2-(2/x)=(-19)
(2/x)^2-(2/x)=(-19)*(1/2)^2-1/2
(2/x)^2-(2/x)=(-19)*(1/4)-1/2
(2/x)^2-(2/x)=(-19)*-.25
(x)^2-(x)=(-4.75)
x^2-X=(-4.75)
X*X-X= (-4.75)
X-x=(-4.75/x)
x= -4.75
2006-09-01 08:39:58 · answer #5 · answered by pzzykat 1 · 0⤊ 0⤋
get a graphing calculator and put each side of that equation in a y= function, the intersection(s) is the answer
2006-09-01 08:21:21 · answer #6 · answered by kirupahost 2 · 0⤊ 0⤋