Question

Solve:
(x/x-1)-(x/x+1)=3+(2x^2/1-x^2)
**((also uploading attachment in comments of actual problem))**

Answer 1

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

Answer 2

Solve for x over the real numbers:
x/(x-1)-x/(x+1) = 3+(2 x^2)/(1-x^2)
Write both sides as a single fraction.
Bring x/(x-1)-x/(x+1) together using the common denominator (x-1) (x+1). Bring 3+(2 x^2)/(1-x^2) together using the common denominator x^2-1:
(2 x)/((x-1) (x+1)) = (x^2-3)/(x^2-1)
Multiply both sides by a polynomial to clear fractions.
Cross multiply:
2 x (x^2-1) = (x-1) (x+1) (x^2-3)
Write the cubic polynomial on the left hand side in standard form.
Expand out terms of the left hand side:
2 x^3-2 x = (x-1) (x+1) (x^2-3)
Write the quartic polynomial on the right hand side in standard form.
Expand out terms of the right hand side:
2 x^3-2 x = x^4-4 x^2+3
Move everything to the left hand side.
Subtract x^4-4 x^2+3 from both sides:
-x^4+2 x^3+4 x^2-2 x-3 = 0
Factor the left hand side.
The left hand side factors into a product with four terms:
-((x-3) (x-1) (x+1)^2) = 0
Multiply both sides by a constant to simplify the equation.
Multiply both sides by -1:
(x-3) (x-1) (x+1)^2 = 0
Solve each term in the product separately.
Split into three equations:
x-3 = 0 or x-1 = 0 or (x+1)^2 = 0
Look at the first equation: Solve for x.
Add 3 to both sides:
x = 3 or x-1 = 0 or (x+1)^2 = 0
Look at the second equation: Solve for x.
Add 1 to both sides:
x = 3 or x = 1 or (x+1)^2 = 0
Look at the third equation: Eliminate the exponent.
Take the square root of both sides:
x = 3 or x = 1 or x+1 = 0
Solve for x.
Subtract 1 from both sides:
x = 3 or x = 1 or x = -1
Now test that these solutions are correct by substituting into the original equation.
Check the solution x = -1.
x/(x-1)-x/(x+1) => -(-1)/(1-1)-1/(-1-1) = infinity^~ ~~ infinity^~
3+(2 x^2)/(1-x^2) => 3+(2 (-1)^2)/(1-(-1)^2) = infinity^~ ~~ infinity^~:
So this solution is incorrect
Check the solution x = 1.
x/(x-1)-x/(x+1) => 1/(1-1)-1/(1+1) = infinity^~ ~~ infinity^~
3+(2 x^2)/(1-x^2) => 3+(2 1^2)/(1-1^2) = infinity^~ ~~ infinity^~:
So this solution is incorrect
Check the solution x = 3.
x/(x-1)-x/(x+1) => 3/(3-1)-3/(1+3) = 3/4
3+(2 x^2)/(1-x^2) => 3+(2 3^2)/(1-3^2) = 3/4:
So this solution is correct
Gather any correct solutions.
The solution is:
Answer: |
| x = 3

Answer 3

there are two answers sorry,x=0,1

Answer 4

since 2x=0
x=0 another 1 is 1-x=0
x=1

Answer 5

whats ur answer?

Answer 6

i didn't know how to do it at all and was getting stuck left and rig.t, i was so confused on how to do it and since im not in school to ask for help i came here..thank you so much for the explanation and help on this one