one of your solutions is extraneous, which means it is not a valid solution to the problem. which solution is extraneous? Explain how you know
I apologize, but I cannot identify which specific solution is extraneous without having access to the problem or solutions provided. However, I can provide you with general information on how to identify an extraneous solution.
An extraneous solution is an apparent solution that does not satisfy the original equation or problem. It may arise when manipulating equations algebraically, particularly when squaring or taking an even root of both sides of the equation. This can introduce additional solutions that do not actually satisfy the original equation.
To identify an extraneous solution, you need to substitute the solution into the original equation and check if it holds true. If the substituted value does not satisfy the equation, it is an extraneous solution.
For example, if the original equation is √(x-1) + 3 = 2, and one solution is x = 1, you would substitute x = 1 back into the equation and see if it holds true:
√(1-1) + 3 = 2
√0 + 3 = 2
0 + 3 = 2
3 = 2
Since 3 is not equal to 2, the solution x = 1 is not valid and, therefore, extraneous.
In summary, without access to specific problem details, it is not possible for me to identify the extraneous solution. However, you can follow the general process explained above to determine the extraneous solution yourself.