Which solution to the equation $\frac{3}{2 g+8}=\frac{g+2}{g^2-16}$ is extraneous?
A. $g=-4$
B. $g=-4$ and $g=16$
C. neither $g=-4$ nor $g=16$
D. $g=16$
Answer (2)
Factor the denominators of the rational expressions.
Solve the equation by clearing the denominators and isolating the variable.
Check the solutions by substituting them back into the original equation to identify any extraneous solutions.
The extraneous solution is g = − 4 .
Explanation
Understanding the Problem We are given the equation 2 g + 8 3 = g 2 − 16 g + 2 and asked to find any extraneous solutions.
Factoring Denominators First, we factor the denominator on the right side of the equation: g 2 − 16 = ( g − 4 ) ( g + 4 ) . We can also factor a 2 out of the denominator on the left side: 2 g + 8 = 2 ( g + 4 ) . Thus, our equation becomes 2 ( g + 4 ) 3 = ( g − 4 ) ( g + 4 ) g + 2 .
Clearing Denominators To solve for g , we multiply both sides of the equation by 2 ( g + 4 ) ( g − 4 ) to eliminate the denominators. This gives us 3 ( g − 4 ) = 2 ( g + 2 ) .
Expanding the Equation Expanding both sides, we have 3 g − 12 = 2 g + 4.
Solving for g Subtracting 2 g from both sides gives g − 12 = 4 . Adding 12 to both sides gives g = 16 .
Checking for Extraneous Solutions Now we need to check for extraneous solutions. Extraneous solutions occur when a solution makes one of the original denominators equal to zero. The original denominators are 2 g + 8 and g 2 − 16 . If g = − 4 , then 2 g + 8 = 2 ( − 4 ) + 8 = 0 and g 2 − 16 = ( − 4 ) 2 − 16 = 0 . Thus, g = − 4 would make both denominators zero, and is an extraneous solution. If g = 16 , then 2 g + 8 = 2 ( 16 ) + 8 = 40 = 0 and g 2 − 16 = ( 16 ) 2 − 16 = 256 − 16 = 240 = 0 . Thus, g = 16 is a valid solution.
Final Answer Therefore, the extraneous solution is g = − 4 .
Examples
When solving equations involving rational expressions, it's crucial to identify extraneous solutions. These are values obtained during the solving process that do not satisfy the original equation, often because they make a denominator zero. For instance, in electrical circuit analysis, you might derive an equation to find the resistance in a circuit. If solving this equation yields a negative resistance value, that would be an extraneous solution since resistance cannot be negative in real-world applications. Similarly, in physics, when calculating projectile motion, an extraneous solution might represent a time before the launch, which is not physically meaningful. Always checking the validity of solutions within the context of the original problem ensures accurate and meaningful results.
The extraneous solution to the equation is g = − 4 , as it makes the denominators zero. The valid solution is g = 16 . Therefore, the answer is option A: g = − 4 .
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