SOLUTION: 2= 3/x-1/x^2 I cant find the restrictions in this problem or the least common faction if you could get that for me then i caould do the rest.

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Question 30347: 2= 3/x-1/x^2 I cant find the restrictions in this problem or the least common faction if you could get that for me then i caould do the rest.
Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website!
2= (3/x)-(1/x^2) ----(1)
Multiply by x^2 through out
2x^2 = 3x-1
2x^2-3x+1 = 0
2x^2-2x-x+1 = 0
(splitting the mid term as the sum of two terms whose product is the product of the square term and the constant term and hence -3x = (-2x)+ (-x) and (-2x)X (-x) = 2x^2 = (2x^2)X1 )
2x(x-1)-1(x-1) = 0
2xp-p= 0 (where p = (x-1) )
p(2x-1) = 0
(x-1)(2x-1) = 0
(x-1) = 0 gives x = 1
(2x-1) = 0 gives 2x = 1 which implies x = 1/2
Answer: x = 1 and x= 1/2
Verification: x = 1 in (1) we get
RHS = 3/x - 1/x^2 = 3/1 -1/1^2 = 3-1 = 2 = LHS
x = 1/2 in (1) we get
RHS = 3/x - 1/x^2 = [3/(1/2)] -1/[(1/2)^2] = 3X2-1/(1/4) = 6-4=2= LHS
Therefore our values are correct.