- Isolate the radical term: \(\sqrt{x-1} = x - 3\).
- Square both sides to get \(x - 1 = x^2 - 6x + 9\), simplifying to \(x^2 - 7x + 10 = 0\).
- Factor the quadratic equation \((x-5)(x-2) = 0\) to find potential solutions \(x=5\) and \(x=2\).
- Verify the solutions against the constraint \(x \geq 3\), identifying \(x=5\) as the only valid solution.
- The final answer is \(\boxed{x=5}\).
Explanation
- Isolating the Radical and Identifying Constraints
To solve the equation \(\sqrt{x-1} - 5 = x - 8\), we first isolate the square root term. By adding 5 to both sides, we get:
\[\sqrt{x-1} = x - 3\]
The square root function \(\sqrt{u}\) (for \(u \geq 0\)) is always non-negative. This implies that the right side of the equation must also be non-negative:
\[x - 3 \geq 0 \implies x \geq 3\]
Additionally, the radicand must be non-negative:
\[x - 1 \geq 0 \implies x \geq 1\]
For both conditions to be satisfied, we must have \(x \geq 3\).
- Squaring Both Sides and Forming a Quadratic Equation
To eliminate the radical, we square both sides of the equation \(\sqrt{x-1} = x - 3\):
\[(\sqrt{x-1})^2 = (x - 3)^2\]
\[x - 1 = x^2 - 6x + 9\]
Rearranging this into a standard quadratic form \(ax^2 + bx + c = 0\):
\[x^2 - 7x + 10 = 0\]
- Solving the Quadratic Equation
We can solve the quadratic equation \(x^2 - 7x + 10 = 0\) by factoring it:
\[(x - 5)(x - 2) = 0\]
This yields two potential solutions: \(x = 5\) and \(x = 2\).
- Verifying Potential Solutions
We must check our potential solutions against the constraint \(x \geq 3\) derived in step 1.
- For \(x = 5\): Since \(5 \geq 3\), this is a valid solution. We can verify this in the original equation: \(\sqrt{5-1} - 5 = 2 - 5 = -3\), and \(5 - 8 = -3\).
- For \(x = 2\): Since \(2 < 3\), this is an extraneous solution. We can verify this in the original equation: \(\sqrt{2-1} - 5 = 1 - 5 = -4\), whereas \(2 - 8 = -6\). Since \(-4
eq -6\), \(x=2\) is not a solution.
Thus, the equation has one real solution, which is \(x=5\).
Examples
This problem demonstrates the importance of checking for extraneous solutions when working with radical equations. In real-life scenarios, such as modeling physical systems where square root relationships appear (e.g., velocity calculations involving kinetic energy \(E = \frac{1}{2}mv^2\) where speed \(v = \sqrt{\frac{2E}{m}}\)), it is crucial to ensure that the derived values conform to the physical constraints of the system (such as speed cannot be negative). Solving these equations correctly ensures that the mathematical model accurately reflects physical reality.