- Rearrange the equation to \(\sqrt{2x-1} = x - 2\).
- Square both sides to obtain \(x^2 - 6x + 5 = 0\).
- Factor the quadratic to find potential solutions \(x=1\) and \(x=5\).
- Verify the roots in the original equation, finding that \(x=5\) is the valid solution and \(x=1\) is extraneous.
- Final answer: \(\boxed{\text{There is only one solution: } x=5. \text{ The solution } x=1 \text{ is an extraneous solution.}}\)
Explanation
- Solving the Algebraic Equation
The given equation is \(\sqrt{2x-1} - x + 2 = 0\). To ensure the equation is well-defined, we must have \(2x - 1 \geq 0\), which implies \(x \geq \frac{1}{2}\). We begin by isolating the radical expression:
\[\sqrt{2x-1} = x - 2\]
Squaring both sides of this equation yields:
\[2x - 1 = (x - 2)^2\]
\[2x - 1 = x^2 - 4x + 4\]
Rearranging the terms to form a quadratic equation, we get:
\[x^2 - 6x + 5 = 0\]
- Verifying Potential Solutions
To solve the quadratic equation \(x^2 - 6x + 5 = 0\), we can factor it:
\[(x - 1)(x - 5) = 0\]
This gives us two potential solutions: \(x = 1\) and \(x = 5\). Since squaring both sides of an equation can introduce extraneous solutions, we must substitute both potential roots back into the original equation \(\sqrt{2x-1} - x + 2 = 0\) to verify them.
For \(x = 1\):
\[\sqrt{2(1) - 1} - 1 + 2 = \sqrt{1} + 1 = 1 + 1 = 2
eq 0\]
Since \(2
eq 0\), \(x = 1\) is an extraneous solution.
For \(x = 5\):
\[\sqrt{2(5) - 1} - 5 + 2 = \sqrt{9} - 3 = 3 - 3 = 0\]
Since \(0 = 0\), \(x = 5\) is the valid solution. Additionally, noting that the isolated form \(\sqrt{2x-1} = x-2\) requires \(x-2 \geq 0\) for the square root to be valid, we confirm \(x \geq 2\), further validating that \(x=5\) is the solution and \(x=1\) is not.
- Conclusion
Based on our verification, the only solution to the given equation is \(x = 5\). The value \(x = 1\) was introduced during the squaring process and does not satisfy the original equation. Thus, the correct statement is that there is only one solution, \(x = 5\), and \(x = 1\) is extraneous.
Examples
This problem illustrates how squaring both sides of an equation can produce extraneous solutions that do not satisfy the original conditions. This is a common situation in various scientific and engineering applications, such as calculating the time taken for a projectile to reach a certain distance or analyzing energy balances where radical terms appear. Ensuring that all mathematical results are verified against the original constraints is a vital step for maintaining accuracy in any analytical field.