Verified Answer :

  • Analyze the domain constraints to ensure the square roots are real and the equation is valid, leading to \(x \geq 2.5\).
  • Square both sides of the equation to eliminate the radical expressions and simplify to a quadratic equation.
  • Factor the resulting quadratic equation to find potential solutions \(x = 25\) and \(x = 1\).
  • Check both values in the original equation to identify the valid solution, yielding \(\boxed{x = 25}\).

Explanation

  1. Domain Analysis and Constraints

We are given the equation \(\sqrt{x} + 2 = \sqrt{2x - 1}\). To solve for \(x\), we first identify the domain of the equation. For the square roots to yield real values, the expressions inside them must be non-negative:

  1. \(x \geq 0\)
  2. \(2x - 1 \geq 0 \implies x \geq \frac{1}{2}\)

Furthermore, since the left-hand side \(\sqrt{x} + 2\) is at least \(2\) (because \(\sqrt{x} \geq 0\)), the right-hand side \(\sqrt{2x - 1}\) must also be at least \(2\). This implies \(\sqrt{2x - 1} \geq 2\), or \(2x - 1 \geq 4\), which simplifies to \(2x \geq 5\) or \(x \geq 2.5\). Thus, any valid solution must satisfy \(x \geq 2.5\).

  1. Algebraic Manipulation

To eliminate the square roots, we square both sides of the original equation:
\[(\sqrt{x} + 2)^2 = (\sqrt{2x - 1})^2\]
Expanding the left side:
\[x + 4\sqrt{x} + 4 = 2x - 1\]
Next, we isolate the term with the square root by subtracting \(x\) and \(4\) from both sides:
\[4\sqrt{x} = 2x - 1 - x - 4\]
\[4\sqrt{x} = x - 5\]

  1. Solving the Quadratic Equation

Now, we square both sides again to eliminate the remaining square root:
\[(4\sqrt{x})^2 = (x - 5)^2\]
\[16x = x^2 - 10x + 25\]
Rearranging the terms into a standard quadratic equation form \(ax^2 + bx + c = 0\):
\[x^2 - 26x + 25 = 0\]
We factor this quadratic equation:
\[(x - 25)(x - 1) = 0\]
This gives us two potential solutions: \(x = 25\) and \(x = 1\).

  1. Verification and Final Conclusion

We must verify the potential solutions against our domain constraint \(x \geq 2.5\) and the original equation.

  1. For \(x = 25\):

\(\sqrt{25} + 2 = 5 + 2 = 7\)
\(\sqrt{2(25) - 1} = \sqrt{50 - 1} = \sqrt{49} = 7\)
Since \(7 = 7\), \(x = 25\) is a valid solution.

  1. For \(x = 1\):

\(\sqrt{1} + 2 = 1 + 2 = 3\)
\(\sqrt{2(1) - 1} = \sqrt{1} = 1\)
Since \(3 eq 1\), \(x = 1\) is an extraneous solution and is invalid.
Thus, the only valid solution is \(x = 25\).

Examples

Understanding how to solve radical equations is crucial in physics and engineering, particularly when dealing with relationships involving velocity or energy. For instance, in problems where energy \(E = \frac{1}{2}mv^2\) relates to speed, you often encounter square roots. Solving these equations correctly ensures accurate predictions of physical quantities, helping engineers design safer systems and calculate precise force requirements.

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