Find all real solutions to the equation (algebra)?

Question

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

Answer 1

Hmmm...

Answer 2

First step is to solve the left hand side with a common denominator as (x-1)(x^2).

Then,  1/(x-1) = (x^2)/(x-1)(x^2) and

          1/(x^2) = (x-1)/(x-1)(x^2)

Next, solve the left hand side as:

   [x^2 + (x-1)] / [(x-1)(x^2)]

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

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

Finally, you place this back to your equation:

   [(x-2)(x+1)] / [(x-1)(x^2)] = 0

In order for this to be equal to 0, either (x-2) = 0 or (x+1) = 0 but make sure (x-1) not = 0 and (x^2) not = 0 (because nothing can be divided by 0).

So, x = 2 or x = -1 given that x != 1 and x != 0

    

Answer 3

Your question --

1/(x-1) + 1/(x^2) = 0 ,  Simplify the left hand side by taking LCM of the denominator.

=>   ( x^2  +  x  -  1 ) / [ x^2 ( x - 1 ) ]  =  0 , Now cross multiply

=>  x^2  +  x  -  1  =  0

  

=>  x  =  (1/2) [  - 1  Ã‚±  Ã¢ÂˆÂš(1+4)  ]  =  (1/2)  [  - 1  Ã‚±  2.236  ]

=>  x  =  + 0.618  or  - 1.618 ........................ Answer

Dr. PKT

Answer 4

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

So,

(x-1) = 0

x = 1

OR

(x^2) = 0

x = 0

So the answer is x = 0,1

Answer 5

does the "/" mean to multiply or divide i forget...:-x