SOLUTION: Find all values of x such that 9(x + 3) = -\frac{8}{x} + \frac{2}{x - 5}. If you find more than one value, then list your solutions, separated by commas.

Question 1209109: Find all values of x such that
9(x + 3) = -\frac{8}{x} + \frac{2}{x - 5}.
If you find more than one value, then list your solutions, separated by commas.
Answer by yurtman(42) (Show Source): You can put this solution on YOUR website!
To solve this equation, we'll first clear the denominators by multiplying both sides by $x(x-5)$:
$$9(x+3)x(x-5) = -8(x-5) + 2x$$
Expand the left side:
$$9x^3 - 45x^2 + 27x^2 - 135x = -8x + 40 + 2x$$
Combine like terms:
$$9x^3 - 18x^2 - 127x - 40 = 0$$
Unfortunately, this is a cubic equation, and there's no simple algebraic method to solve it directly. We can use numerical methods or software tools to find approximate solutions.
However, we can try to factor the equation to see if we can find any rational solutions. By using the Rational Root Theorem, we can test potential rational roots.
By trial and error, we find that $x = -1$ is a solution. This means that $(x+1)$ is a factor of the cubic polynomial. We can perform polynomial long division to find the other factor:
$$9x^3 - 18x^2 - 127x - 40 = (x+1)(9x^2 - 27x - 40)$$
Now, we can solve the quadratic equation $9x^2 - 27x - 40 = 0$ using the quadratic formula:
$$x = \frac{-(-27) \pm \sqrt{(-27)^2 - 4(9)(-40)}}{2(9)}$$
Simplifying:
$$x = \frac{27 \pm \sqrt{2001}}{18}$$
So, the solutions to the original equation are:
$$x = -1, \frac{27 + \sqrt{2001}}{18}, \frac{27 - \sqrt{2001}}{18}$$

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