2006-07-25 18:51:13 · 9 answers · asked by Cowman 1 in Science & Mathematics ➔ Mathematics
((x^2)/(x + 1)) - 1 = (3x)/(x + 1)
((x^2)/(x + 1)) - ((x + 1)/(x + 1)) = (3x)/(x + 1)
(x^2 - (x + 1))/(x + 1) = (3x)/(x + 1)
(x^2 - x - 1)/(x + 1) = (3x)/(x + 1)
x^2 - x - 1 = 3x
x^2 - 4x - 1 = 0
x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-(-4) ± sqrt((-4)^2 - 4(1)(-1)))/(2(1))
x = (4 ± sqrt(16 + 4))/2
x = (4 ± sqrt(20))/2
x = (4 ± sqrt(4 * 5))/2
x = (4 ± 2sqrt(5))/2
ANS : x = 2 ± sqrt(5)
2006-07-26 03:01:01 · answer #1 · answered by Sherman81 6 · 1⤊ 0⤋
X=0
2006-07-25 18:56:09 · answer #2 · answered by li_da_mi 3 · 0⤊ 0⤋
x^2/(x+1)-1 = 3x/x-1
x^2/x = 3x/x-1 The 1 -1 drops out
x * (x^2/x) = x * (3x/x+1) Multiply both sides by x
x^2 = 3x^2/x+1 The result
(1/x^2)x^2 = (3x^2/x+1)(1/x^2) multiply both side by 1/x^2
1= 3x^2/ ((x+1)(x^2)) The result
x+1 = 3x^2/x^2 Multiply both sides by x+1
x+1 = 3 Cancel out he x^2
x = 3 -1
x=2
2006-07-25 19:15:04 · answer #3 · answered by Anonymous · 0⤊ 0⤋
X^2/(X+1)-1 = 3X/X-1
X^2/X = 3X/X-1
X * (X^2/X) = X * (3X/X+1) Multiply both sides by X
X^2 = 3X^2/X+1
(1/X^2)X^2 = (3X^2/X+1)(1/X^2) multiply both side by 1/X^2
1= 3X^2/ ((X+1)(X^2))
X+1 = 3X^2/X^2 Multiply both sides by X+1
X+1 = 3 Cancel out he X^2
X = 3 -1
x=2
2006-07-25 20:53:32 · answer #4 · answered by Jatta 2 · 0⤊ 0⤋
[x^2/(x+1)]-1 = 3x/(x+1)
x^2-4x-1 = 0
x = [4-sqrt(20)]/2, or
x = [4+sqrt(20)]/2
2006-07-25 19:01:56 · answer #5 · answered by mekaban 3 · 0⤊ 0⤋
...Cowman, stop cheating.. Your never going to learn by people giving you the answer.
...What if these people have recognized what you're doing and given you any number and variable to make it look good?
...Stop being so inquisitive and try to understand. Math.com is perfect for you...
Enjoy...
--Rob :)~
2006-07-25 19:33:32 · answer #6 · answered by stealth_n700ms 4 · 0⤊ 0⤋
(x^2)/(x+1)-1=(3x)/(x+1)
(x^2)/(x+1)-(3x)/(x+1)=1
(x^2-3x)/(x+1)=1
(x^2-3x)=x+1
x^2-4x-1=0
using -b+-sqrt(b^2-4ac)/2a
x=4.23606797749979
or
x=-0.2360679774997898
2006-07-25 19:03:33 · answer #7 · answered by Anonymous · 0⤊ 0⤋
try to write more mathematically in future.
2006-07-25 20:12:58 · answer #8 · answered by rajesh bhowmick 2 · 0⤊ 0⤋
square root (5) +/- 0.5
2006-07-25 19:03:21 · answer #9 · answered by VP 2 · 0⤊ 0⤋