Solve the equation for x: 1/x-5 = 1/x+3 + x-1/x^2-2x-15

1 answer

To solve the equation

\[ \frac{1}{x} - 5 = \frac{1}{x + 3} + \frac{x - 1}{x^2 - 2x - 15}, \]

we first factor the quadratic in the denominator on the right side. The expression \( x^2 - 2x - 15 \) factors to

\[ (x - 5)(x + 3). \]

Therefore, we can rewrite the equation as

\[ \frac{1}{x} - 5 = \frac{1}{x + 3} + \frac{x - 1}{(x - 5)(x + 3)}. \]

Next, we multiply both sides of the equation by \( x(x + 3)(x - 5) \) to eliminate the denominators. This gives us:

\[ (x - 5)(x)(x + 3) \left(\frac{1}{x} - 5\right) = (x - 5)(x) + (x - 1)(x). \]

The left side becomes:

\[ (x - 5)(x + 3) - 5x(x - 5)(x + 3). \]

Distributing the terms:

\[ (x^2 - 2x - 15) - 5x^2 + 25x, \] which simplifies to: \[ -4x^2 + 23x - 15. \]

Reorganizing gives us:

\[ -4x^2 + 23x - 15 = x^2 - 5x. \]

Now, moving all terms to one side:

\[ -4x^2 + 23x - 15 - x^2 + 5x = 0. \] This simplifies to:

\[ -5x^2 + 28x - 15 = 0. \]

Multiplying through by -1 gives:

\[ 5x^2 - 28x + 15 = 0. \]

Next, we apply the quadratic formula where \( a = 5 \), \( b = -28 \), and \( c = 15 \):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 5 \cdot 15}}{2 \cdot 5}. \]

Calculating the discriminant:

\[ 28^2 - 4 \cdot 5 \cdot 15 = 784 - 300 = 484. \]

Now substituting this back into the quadratic formula:

\[ x = \frac{28 \pm \sqrt{484}}{10} = \frac{28 \pm 22}{10}. \]

Calculating the two cases:

  1. For the positive root: \[ x = \frac{28 + 22}{10} = \frac{50}{10} = 5. \]

  2. For the negative root: \[ x = \frac{28 - 22}{10} = \frac{6}{10} = \frac{3}{5}. \]

Therefore, the potential solutions are \( x = 5 \) and \( x = \frac{3}{5} \).

Next, we need to check for restrictions from the original equation denominators, specifically:

  1. \( x \neq 0 \)
  2. \( x + 3 \neq 0 \), thus \( x \neq -3 \)
  3. \( x - 5 \neq 0 \), thus \( x \neq 5 \)

Since \( x = 5 \) results in a division by zero, it is not a valid solution.

Thus, the only valid solution is

\[ \boxed{\frac{3}{5}}. \]