How do I solve 2 /(x-1)= 3(x^2-1) + 1?

Answer 1

2 /(x-1)= 3(x^2-1) + 1

2 = 3(x^2-1) + 1 * (x-1)

2 = [3x^2 - 3 + 1] * (x-1)

2 = (3x^2 - 2) * (x-1)

0 = 3x^3 - 3x^2 - 2x + 2 - 2

0 = 3x^3 - 3x^2 - 2x
0 = x ( 3x^2 - 3x - 2)

x = 0 and 0 = 3x^2 - 3x - 2

Use quadratic equation after this:
x = [-b +/- √(b^2 - 4ac)] / 2a

x = [3 +/- √(9-4(3)(-2)] / 2(3)
x = [3 +/- √(9+24)] / 6

x = [3 +/- √(33)] / 6

3 answers: x = 0, x = [3 +/- √(33)] / 6

Answer 2

2 /(x-1) = 3(x^2-1) + 1
2 /(x-1) = 3(x-1)(x+1) + 1
(x-1) * 2/(x-1) = 3(x-1)(x+1)(x-1) + 1(x-1)
2 = 3(x^2-1)(x-1) + (x - 1)
2 = 3(x^3 - x^2 - x + 1) + x - 1
2 = 3x^3 - 3x^2 - 3x + 3 + x - 1
0 = 3x^3 - 3x^2 - 2x

then solve for x using cubic formula
x = 1.457, x = 0, x = - 0.457 (approx)

Answer 3

2 /(x-1)= 3(x^2-1) + 1
2 = (3x^2 - 3 + 1)(x-1)
(3x^2 - 2)(x-1) - 2 = 0
3x^3 - 2x - 3x^2 + 2 - 2 = 0
3x^3 - 2x - 3x^2 = 0
x(3x^2 - 2 - 3x) = 0
x(3x^2 - 3x - 2) = 0
=> x = 0 and 3x^2 - 3x - 2 = 0

For 3x^2 - 3x - 2 = 0 the solution for x can be obtained by using the formula [-b +/- sqrt(b^2 - 4ac)]/2a

x = [3 +/- sqrt(9 + 24)]/6
x = [3 +/- sqrt(33)]/6

Answer 4

2 / (x - 1) = 3x² - 3 + 1
2 / (x - 1) = 3x² - 2
2 = (3x² - 2) (x - 1)
2 = 3x³ - 3x² - 2x + 2
3x³ - 3x² - 2x = 0
x (3x² - 3x - 2) = 0
x = 0 or
3x² - 3x - 2 = 0
x = [ 3 ± √(9 + 24)] / 6
x = [ 3 ± √(33)] / 6

Answer 5

Through algebra!
Multiply out the brackets, then multiply through by x-1.
Graph the resultant cubic equation, and find where it cuts the axis.

Answer 6

Easy calculator.