Find all the real solutions to
\[\frac{x^2 + 4x}{x - 1} + \frac{72x - 72}{x^2 + 4x} - 18 = 0.\]
Enter all the solutions, separated by commas.
+128656 \(\frac{x^2 + 4x}{x - 1} + \frac{72x - 72}{x^2 + 4x} - 18 = 0.\)
Let u = x^2 + 4x
_________
x - 1
And note that 72 * 1 72 ( x - 1) 72x - 72
___ = __________ = _________
u x^2 + 4x x^2 + 4x
So we have
u + 72 / u - 18 = 0 multiply through by u
u^2 + 72 - 18 u = 0 rearrange as
u^2 - 18u + 72 = 0 factor
( u - 12) ( u - 6) = 0
Setting each factor to 0 and solving for u we have that
u = 6 or u = 12
But we need to solve for x.......so we have that
x^2 + 4x
_______ = 6
x - 1
x^2 + 4x = 6 ( x - 1)
x^2 + 4x - 6x + 1 = 0
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1
Reject this as it makes an original denomiator = 0
Also
x^2 + 4x
_______ = 12
x - 1
x^2 + 4x = 12 (x - 1)
x^2 + 4x -12x + 12 = 0
x^2 - 8x + 12 = 0
(x - 2) ( x - 6) = 0
Setting each factor to 0 and solve for x and we get the two real solutions
x = 2 or x = 6
