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Difficulty: Very hardRadical and Rational Equations

What is the sum of all real solutions to the equation x−5x−1=2x+1−8x2−1\frac{x - 5}{x - 1} = \frac{2}{x + 1} - \frac{8}{x^2 - 1}?

  1. A
    1
  2. 5Answer
  3. C
    6
  4. D
    -6

Answer

The sum of all real solutions to the equation is 55.
Multiplying the equation by the least common denominator, (x−1)(x+1)(x - 1)(x + 1), yields (x−5)(x+1)=2(x−1)−8(x - 5)(x + 1) = 2(x - 1) - 8. Expanding and simplifying leads to the quadratic equation x2−6x+5=0x^2 - 6x + 5 = 0, which factors as (x−1)(x−5)=0(x - 1)(x - 5) = 0. This gives potential solutions of x=1x = 1 and x=5x = 5. Since x=1x = 1 makes the denominators of the original terms zero, it is extraneous and must be discarded. The only valid solution is x=5x = 5, and therefore the sum of all real solutions is 55.

Step-by-Step Solution

1
Find the least common denominator of the rational terms.
The denominators are x−1x - 1, x+1x + 1, and x2−1x^2 - 1. Since x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1), the least common denominator is (x−1)(x+1)(x - 1)(x + 1).
Finding the least common denominator allows us to eliminate the fractions by multiplying both sides.
2
Multiply the entire equation by the least common denominator (x−1)(x+1)(x - 1)(x + 1), assuming x≠1x \neq 1 and x≠−1x \neq -1.
(x−5)(x+1)=2(x−1)−8(x - 5)(x + 1) = 2(x - 1) - 8
This clears all rational expressions, leaving a polynomial equation.
3
Expand both sides of the equation and combine like terms.
x2−4x−5=2x−2−8⇒x2−4x−5=2x−10x^2 - 4x - 5 = 2x - 2 - 8 \Rightarrow x^2 - 4x - 5 = 2x - 10
Expanding allows the simplification of terms on each side of the equation.
4
Rearrange the equation to set it equal to zero.
x2−6x+5=0x^2 - 6x + 5 = 0
Moving all terms to one side forms a standard quadratic equation which can then be solved.
5
Factor the quadratic equation.
(x−1)(x−5)=0(x - 1)(x - 5) = 0, which gives potential solutions x=1x = 1 and x=5x = 5.
Factoring allows us to find the roots of the quadratic equation.
6
Check the potential solutions in the original equation to identify any extraneous solutions.
Substituting x=1x = 1 results in division by zero in the terms x−5x−1\frac{x - 5}{x - 1} and 8x2−1\frac{8}{x^2 - 1}, so x=1x = 1 is extraneous. Substituting x=5x = 5 yields 0=00 = 0, meaning x=5x = 5 is a valid real solution.
Multiplying by variables can introduce extraneous solutions that make the original denominators zero.
7
Calculate the sum of all valid solutions.
Since x=5x = 5 is the only valid solution, the sum of all real solutions is 55.
The question asks for the sum of all real solutions.

Key Concept

Solving rational equations by clearing denominators and checking for extraneous solutions.
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Topics

Radical and Rational Equations

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