How do you solve # x^2/(x^2-4) = x/(x+2)-2/(2-x)#?

Question

Nathan Axel · Accepted Answer

There is no solution

#x^2/(x^2-4)=x/(x+2)-2/(2-x)# Becomes #x^2/(x^2-4)=x/(x+2)+2/(x-2)# On the right side, multiply and divide first fraction with #x-2# On the right side, multiply and divide second fraction with #x+2# We get, Becomes #x^2/(x^2-4)=(x(x-2))/((x+2)(x-2))+(2(x+2))/((x-2)(x+2))# Becomes #x^2/(x^2-4)=(x^2-2x + 2x + 4)/(x^2-4)# Becomes #x^2/(x^2-4)=(x^2 + 4)/(x^2-4)# Becomes #x^2=(x^2 + 4)# There's not a fix.

Ethan Adkins · Answer

undefined To solve the equation x^2/(x^2-4) = x/(x+2) - 2/(2-x), we can follow these steps:

1. Start by finding a common denominator for the fractions on the right side of the equation. The common denominator is (x+2)(2-x).

2. Rewrite the fractions on the right side with the common denominator:
   x/(x+2) = x(2-x)/[(x+2)(2-x)]
   -2/(2-x) = -2(x+2)/[(x+2)(2-x)]

3. Simplify the fractions on the right side:
   x/(x+2) = (2x-x^2)/[(x+2)(2-x)]
   -2/(2-x) = -2(x+2)/[(x+2)(2-x)]

4. Combine the fractions on the right side:
   (2x-x^2)/[(x+2)(2-x)] - 2(x+2)/[(x+2)(2-x)]

5. Now, we can combine the fractions on the right side into a single fraction:
   [(2x-x^2) - 2(x+2)]/[(x+2)(2-x)]

6. Simplify the numerator:
   (2x-x^2 - 2x - 4)/[(x+2)(2-x)]

7. Combine like terms in the numerator:
   (-x^2 - 4)/[(x+2)(2-x)]

8. Factor the numerator:
   -(x^2 + 4)/[(x+2)(2-x)]

9. Since the equation is set equal to zero, we can set the numerator equal to zero:
   x^2 + 4 = 0

10. Solve for x by taking the square root of both sides:
   x = ±√(-4)

11. Since the square root of a negative number is not a real number, there are no real solutions to the equation.

Therefore, the equation x^2/(x^2-4) = x/(x+2) - 2/(2-x) has no real solutions.