Question

Which statement best reflects the solution(s) of the equation?

2x−1−−−−−√−x+2=0



Responses

There is only one solution: x = 5.
The solution x = 1 is an extraneous solution.

There is only one solution: x = 5. The solution x = 1 is an extraneous solution.

There is only one solution: x = 5.
The solution x = 0 is an extraneous solution.

There is only one solution: x = 5. The solution x = 0 is an extraneous solution.

There are two solutions: x = 1 and x = 5.
There are two solutions: x = 1 and x = 5.

There is only one solution: x = 1.
The solution x = 5 is an extraneous solution.

Answer 1

To determine the solution(s) of the equation \( \sqrt{2x - 1} - x + 2 = 0 \), we first isolate the square root:

\[
\sqrt{2x - 1} = x - 2
\]

Next, we square both sides to eliminate the square root:

\[
2x - 1 = (x - 2)^2
\]

Expanding the right side gives:

\[
2x - 1 = x^2 - 4x + 4
\]

Now, rearranging the equation leads to:

\[
0 = x^2 - 4x - 2x + 4 + 1
\]
\[
0 = x^2 - 6x + 5
\]

Next, we can factor the quadratic:

\[
0 = (x - 1)(x - 5)
\]

Setting each factor to zero gives the potential solutions:

\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\]

We need to check if either solution is extraneous by substituting back into the original equation:

1. For \( x = 1 \):
\[
\sqrt{2(1) - 1} - 1 + 2 = \sqrt{2 - 1} - 1 + 2 = \sqrt{1} - 1 + 2 = 1 - 1 + 2 = 2 \neq 0
\]
Therefore, \( x = 1 \) is an extraneous solution.

2. For \( x = 5 \):
\[
\sqrt{2(5) - 1} - 5 + 2 = \sqrt{10 - 1} - 5 + 2 = \sqrt{9} - 5 + 2 = 3 - 5 + 2 = 0
\]
Therefore, \( x = 5 \) is a valid solution.

Based on this analysis, we conclude that:

- There is only one valid solution: \( x = 5 \).
- The solution \( x = 1 \) is an extraneous solution.

Thus, the best statement reflecting the solution(s) of the equation is:

"There is only one solution: x = 5. The solution x = 1 is an extraneous solution."