What is the solution to the equation?
$(x-2)^{\frac{1}{2}}+4=x$
We are asked to solve the equation \((x-2)^{\frac{1}{2}} + 4 = x\). First, note that the expression \((x-2)^{\frac{1}{2}}\) is equivalent to \(\sqrt{x-2}\). For this square root to be a real number, we must have \(x-2 \geq 0\), so \(x \geq 2\). Additionally, isolating the radical gives \(\sqrt{x-2} = x-4\). Since the square root is non-negative, we must also have \(x-4 \geq 0\), which means \(x \geq 4\). Any potential solution must satisfy \(x \geq 4\).
To solve \(\sqrt{x-2} = x-4\), we square both sides of the equation:
\[(\sqrt{x-2})^2 = (x-4)^2\]
This simplifies to:
\[x - 2 = x^2 - 8x + 16\]
Rearranging all terms to one side, we obtain a quadratic equation:
\[x^2 - 9x + 18 = 0\]
We can factor this quadratic by looking for two numbers that multiply to 18 and add to -9. These numbers are -6 and -3. Thus:
\[(x-6)(x-3) = 0\]
The potential solutions are \(x = 6\) and \(x = 3\).
Now, we must verify our candidates against the original equation and our domain constraint \(x \geq 4\).
Therefore, the only valid solution is \(x = 6\). Comparing this to the given options, the correct answer is \(6\).
Equations involving square roots often arise when calculating the time it takes for an object to reach a certain distance under acceleration, or when finding dimensions in geometry where the diagonal length is known. By learning to handle extraneous solutions, one gains the skill to verify findings against physical constraints, such as ensuring that time or distance values remain positive in real-world applications. This process is fundamental in engineering and physics, where mathematical results must always align with the constraints of the physical problem.