What is true about the solution of $\frac{x^2}{2 x-6}=\frac{9}{6 x-18}$?
A. $x= \pm \sqrt{3}$, and they are actual solutions.
B. $x= \pm \sqrt{3}$, but they are extraneous solutions.
C. $x=3$, and it is an actual solution.
D. $x=-3$, but it is an extraneous solution.
Answer (2)
Factor the denominators of the rational expressions.
Simplify the equation by multiplying both sides by a common factor, noting any restrictions on x .
Solve the resulting equation for x , obtaining x = ± 3 .
Verify that these solutions are not extraneous by checking them in the original equation, confirming that x = ± 3 are actual solutions.
The solutions are x = ± 3 , and they are actual solutions. x = ± 3
Explanation
Problem Analysis We are given the equation 2 x − 6 x 2 = 6 x − 18 9 and we need to determine the nature of its solutions.
Factoring and Simplifying First, let's factor the denominators: 2 x − 6 = 2 ( x − 3 ) and 6 x − 18 = 6 ( x − 3 ) .
So the equation becomes 2 ( x − 3 ) x 2 = 6 ( x − 3 ) 9 . We can simplify this equation by multiplying both sides by 6 ( x − 3 ) , but we must keep in mind that x = 3 , since that would make the denominators zero.
Solving for x Multiplying both sides by 6 ( x − 3 ) (assuming x = 3 ) gives: 6 ( x − 3 ) ⋅ 2 ( x − 3 ) x 2 = 6 ( x − 3 ) ⋅ 6 ( x − 3 ) 9 3 x 2 = 9. Dividing both sides by 3, we get x 2 = 3. Taking the square root of both sides, we find x = ± 3 .
Checking for Extraneous Solutions Now we need to check if these solutions are valid. We already know that x cannot be 3, so we need to check if x = 3 and x = − 3 are solutions to the original equation. If x = 3 , the left side is 2 ( 3 ) − 6 ( 3 ) 2 = 2 3 − 6 3 . The right side is 6 ( 3 ) − 18 9 = 6 3 − 18 9 = 2 3 − 6 3 . Since the left side equals the right side, x = 3 is a solution. If x = − 3 , the left side is 2 ( − 3 ) − 6 ( − 3 ) 2 = − 2 3 − 6 3 . The right side is 6 ( − 3 ) − 18 9 = − 6 3 − 18 9 = − 2 3 − 6 3 . Since the left side equals the right side, x = − 3 is a solution.
Conclusion Therefore, the solutions are x = 3 and x = − 3 , and they are actual solutions.
Examples
When solving equations involving rational expressions, it's crucial to identify and exclude values that make the denominator zero. These values are called extraneous solutions because they satisfy the simplified equation but not the original one. For instance, consider designing a bridge where the load distribution is modeled by a rational function. We must ensure that the design parameters (analogous to 'x' in our equation) do not lead to undefined stress values (where the denominator is zero), as this could compromise the bridge's structural integrity. Understanding extraneous solutions helps engineers avoid such critical design flaws.
The solutions of the equation 2 x − 6 x 2 = 6 x − 18 9 are x = ± 3 , and both solutions are actual solutions. Therefore, the correct answer is option A.
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