X/x+1 + 12/x+1 = 2/x^2 - 1?

Question

/ = fraction

x^2 = x to the second

any other questions, please ask!

Answer 1

x/(x + 1) + 12/(x + 1) = 2/(x^2 - 1)

x/(x + 1) + 12/(x + 1) = 2/(x + 1)(x - 1)

[x + 1][x - 1][x/(x + 1) + 12/(x + 1)] = [x + 1][x - 1][2/(x + 1)(x - 1)]

x[x - 1] + 12[x - 1] = 2

x^2 - x + 12x - 12 = 2

x^2 + 11x - 12 - 2 = 0

x^2 + 11x - 14 = 0

x = [-b ±√(b^2 - 4ac)]/2a

a = 1

b = 11

c = -14

x = [-11 ±√(121 + 56)]/2

x = [-11 ±√177]/2

x ≈ [-11 ±13.3]/2

x ≈ [-11 + 13.3]/2

x ≈ 2.3/2

x ≈ 1.15

x ≈ [-11 - 13.3]/2

x ≈ -24.3/2

x ≈ -12.15

∴ x ≈ -12.15 , 1.15

Answer 2

SHOULD be written as :-

x / (x + 1) + 12 / (x + 1) = 2 / (x² - 1)

x / (x + 1) + 12 / (x + 1) = 2 / (x - 1)(x + 1)

x (x - 1) + 12 (x - 1) = 2

x² - x + 12x - 12 = 2

x² + 11x - 14 = 0

x = [ - 11 ± √ (121 + 56 ) ] / 2

x = [ - 11 ± √ (177 ) ] / 2  

Answer 3

x/x+1 + 12/x+1 = 2/x^2 - 1

x/x+1 + 12/x+1 = 2/(x-1) (x+1)

divide all terms by x+1

x + 12 = 2/x-1

multiply both sides by x-1

(x-1) (x + 12) = 2

x^2+11x-12 = 2

x^2+11x-14 = 0

Answer 4

x/x+1 + 12/x+1 = 2/(x-1)(x+1), (x+12)/(x+1) =2/(x+1)(x-1),assume

x=-1then Infinitive=infinitive.Then x+12=2/x-1,x^2+11x-12=2

x^2+11x-14=0,x=(-11-13.304^1/2)/2 and (-11+13.304)/2