Find all solutions of the equation?

Question

Can someone walk me through this problem?
1/x - 1/(x+1) = 3

I started by multiplying the whole equation by x(x+1), so I could find the common denominator. I got:
x + 1 - x = 3x(x+1) or
1 = 3x^2 + 3x

Now I'm stuck. Did I do this first part right, and if so, where do I take it from here?

Answer 1

3x^2 + 3x - 1 = 0

now use the quadratic formula, to find the values of x that satisfies the quadratic equation,
x = [-b +/- (b^2 - 4ac)^1/2]/2a
where, a = 3, b = 3 and c = -1

x = [-3 +/- (9 + 12)^1/2 ]/6
x = [-3 +/- (21)^1/2 ]/6

Answer 2

Solve it as a quadratic equation:

3x^2 + 3x - 1 = 0

x1,2 = [-3 +- sqrt(3^2 - 4*3*(-1))] / 6 =
= [-3 +- sqrt (9 + 12]/6 = [-3 +- sqrt(21)] / 6 =
= -0.5 +- sqrt(21)/6

x1 = -0.5 + sqrt(21)/6
x2 = -0.5 - sqrt(21)/6

Don't worry, none of them is 0 or -1 (because the denominators in the equation are x and x+1)

Answer 3

You did it right. OK so from 1 = 3x^2 + 3x you have to subtract the one to come up with 3x^2 + 3x - 1= 0

from here you got to use the quadratic formula
x= -b plus or minus the square root of b^2 - 4(a)(c) all over 2(a)
where a= 3 b= 3 and c= -1

you should get something like -3 plus or minus the square root of (3)^2 -4(3)(-1) all over 2(3)..

= -3 plus or minus the square root of 9+12 over 6

= -3 plus or minus the square root of 21 over 6

so your roots should be -3 plus the square root of 21 over 6

and -3 minus the square root of 21 over 6

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