What is the solution of $x=2+\sqrt{x-2}$ ?
The given equation is \(x = 2 + \sqrt{x - 2}\). First, we identify the domain of the equation. Since the expression under the square root must be non-negative, we require \(x - 2 \geq 0\), which implies \(x \geq 2\). This is our domain constraint for any potential solution.
To solve the equation \(x = 2 + \sqrt{x - 2}\), we first isolate the radical term. By subtracting 2 from both sides, we get:
\[\sqrt{x - 2} = x - 2.\]
To eliminate the square root, we square both sides of the equation, ensuring that we keep the condition \(x - 2 \geq 0\) in mind:
\[(\sqrt{x - 2})^2 = (x - 2)^2.\]
This simplifies to:
\[x - 2 = (x - 2)^2.\]
Expanding the right side:
\[x - 2 = x^2 - 4x + 4.\]
Rearranging all terms to one side to form a quadratic equation:
\[x^2 - 5x + 6 = 0.\]
The quadratic equation \(x^2 - 5x + 6 = 0\) can be factored by finding two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. Thus, we have:
\[(x - 2)(x - 3) = 0.\]
This gives two potential solutions for \(x\):
\[x = 2 \quad \text{or} \quad x = 3.\]
Finally, we check both solutions against the original equation and our domain constraint (\(x \geq 2\)).
For \(x = 2\): \(2 = 2 + \sqrt{2 - 2}\) becomes \(2 = 2 + 0\), which is correct.
For \(x = 3\): \(3 = 2 + \sqrt{3 - 2}\) becomes \(3 = 2 + 1\), which is correct.
Both solutions satisfy the original equation.
This problem demonstrates how to solve equations involving radicals, a skill essential in fields like engineering and physics. For instance, when calculating the velocity of an object or the tension in a cable, radical expressions often appear. By squaring both sides and solving the resulting quadratic equation, you can determine the specific points where physical parameters balance, ensuring your designs or models are accurate.