Question

cphs : advanced algebra: concepts and connections - block (27.0831030) rational equations which solution to the equation \\( \frac{1}{x - 1} = \frac{x - 2}{2x^2 - 2} \\) is extraneous? options: \\( x = -4 \\), \\( x = 1 \\), \\( x = 1 \\) and \\( x = -4 \\), neither \\( x = 1 \\) or \\( x = -4 \\)

Explanation:

Step1: Identify the equation

The equation is \(\frac{1}{x - 1}=\frac{x - 2}{2x^2 - 2}\). First, factor the denominators. The denominator \(2x^2 - 2\) can be factored as \(2(x^2 - 1)=2(x - 1)(x + 1)\) (using the difference of squares \(a^2 - b^2=(a - b)(a + b)\)). The domain of the original equation excludes \(x = 1\) and \(x=-1\) (since these make the denominators zero).

Step2: Simplify the equation

Multiply both sides by the least common denominator (LCD), which is \(2(x - 1)(x + 1)\) (for \(x
eq1,-1\)):

$$ 2(x - 1)(x + 1)\cdot\frac{1}{x - 1}=2(x - 1)(x + 1)\cdot\frac{x - 2}{2(x - 1)(x + 1)} $$

Simplify both sides:

• Left side: The \((x - 1)\) terms cancel, leaving \(2(x + 1)\).
• Right side: The \(2(x - 1)(x + 1)\) terms cancel, leaving \((x - 2)\).

So the simplified equation is \(2(x + 1)=x - 2\).

Step3: Solve the simplified equation

Expand the left side: \(2x+2=x - 2\).
Subtract \(x\) from both sides: \(x + 2=-2\).
Subtract 2 from both sides: \(x=-4\).

Step4: Check for extraneous solutions

Now, check if \(x = -4\) is in the domain (it is, since \(-4
eq1\) and \(-4
eq - 1\)). Now, check the other "solution" \(x = 1\): but \(x = 1\) makes the original denominator \(x - 1\) zero, so \(x = 1\) is not in the domain and is extraneous. Let's verify \(x=-4\) in the original equation:

Left side: \(\frac{1}{-4 - 1}=\frac{1}{-5}=-\frac{1}{5}\).

Right side: \(\frac{-4 - 2}{2(-4)^2 - 2}=\frac{-6}{2(16)-2}=\frac{-6}{32 - 2}=\frac{-6}{30}=-\frac{1}{5}\). So \(x=-4\) is a valid solution, and \(x = 1\) is extraneous.

Answer:

\(x = 1\)