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Difficulty: Very hardRadical and Rational Equations
If xx is a real solution to the equation
2x−5x−3+x−5x2−4x+3=−1x−1\frac{2x - 5}{x - 3} + \frac{x - 5}{x^2 - 4x + 3} = -\frac{1}{x - 1}
what is the value of 1−2x1 - 2x?
  1. A
    -5
  2. 2Answer
  3. C
    -2
  4. D
    0

Answer

2
The correct answer is 2. Multiplying the entire equation by the least common denominator (x−3)(x−1)(x - 3)(x - 1) simplifies the rational equation to the quadratic equation 2x2−5x−3=02x^2 - 5x - 3 = 0. This quadratic factors into (2x+1)(x−3)=0(2x + 1)(x - 3) = 0, yielding two potential solutions: x=−1/2x = -1/2 and x=3x = 3. However, x=3x = 3 is an extraneous solution because it results in a denominator of zero in the original equation. Thus, the only valid real solution is x=−1/2x = -1/2. Substituting this into the expression 1−2x1 - 2x yields 1−2(−1/2)=21 - 2(-1/2) = 2.

Step-by-Step Solution

1
Factor the quadratic expression in the denominator to identify the least common denominator.
The quadratic in the denominator of the second term factors as x2−4x+3=(x−3)(x−1)x^2 - 4x + 3 = (x - 3)(x - 1). Thus, the least common denominator for all terms in the equation is (x−3)(x−1)(x - 3)(x - 1), with the restrictions that x≠3x \neq 3 and x≠1x \neq 1.
Finding a common denominator allows us to multiply both sides of the equation to clear all rational expressions.
2
Multiply both sides of the equation by the least common denominator (x−3)(x−1)(x - 3)(x - 1).
(2x−5)(x−1)+(x−5)=−1(x−3)(2x - 5)(x - 1) + (x - 5) = -1(x - 3)
This clears the fractions and converts the rational equation into a polynomial equation.
3
Expand and simplify both sides of the equation.
(2x2−7x+5)+(x−5)=−x+3(2x^2 - 7x + 5) + (x - 5) = -x + 3
2x2−6x=−x+32x^2 - 6x = -x + 3
Expanding the terms allows us to combine like terms and set up a quadratic equation.
4
Rearrange the terms to set the quadratic equation equal to zero.
2x2−5x−3=02x^2 - 5x - 3 = 0
Quadratic equations must be set to zero to be solved by factoring or using the quadratic formula.
5
Solve the quadratic equation by factoring.
(2x+1)(x−3)=0(2x + 1)(x - 3) = 0
This gives two potential solutions: x=−1/2x = -1/2 and x=3x = 3.
Factoring is the most direct algebraic method to find the roots of the quadratic equation.
6
Check the potential solutions against the original domain restrictions to identify any extraneous solutions.
For x=3x = 3, the denominators x−3x - 3 and x2−4x+3x^2 - 4x + 3 equal zero, so x=3x = 3 is an extraneous solution. For x=−1/2x = -1/2, the denominators are non-zero, so x=−1/2x = -1/2 is the only valid solution.
Squaring or multiplying by variable expressions can introduce solutions that are undefined in the original equation.
7
Substitute the valid solution x=−1/2x = -1/2 into the expression 1−2x1 - 2x.
1−2(−12)=1+1=21 - 2\left(-\frac{1}{2}\right) = 1 + 1 = 2
The question asks for the value of the expression 1−2x1 - 2x, not the value of xx itself.

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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