Solve the square root equation $\sqrt{2 x-3}=5$.
The given equation is \(\sqrt{2x-3} = 5\). To find the value of \(x\), we first identify that the square root is defined for non-negative values. Thus, the expression inside the radical must satisfy \(2x-3 \ge 0\), which means \(x \ge 1.5\). Our objective is to isolate \(x\) and confirm the validity of the result.
To eliminate the square root, we square both sides of the equation:
\[(\sqrt{2x-3})^2 = 5^2\]
This simplifies to:
\[2x - 3 = 25\]
Now, we solve for \(x\) by performing algebraic operations. Add 3 to both sides:
\[2x = 25 + 3\]
\[2x = 28\]
Divide both sides by 2:
\[x = \frac{28}{2}\]
\[x = 14\]
Finally, verify the solution by substituting \(x = 14\) back into the original equation:
\[\sqrt{2(14) - 3} = \sqrt{28 - 3} = \sqrt{25} = 5\]
The solution satisfies the original equation and the condition \(x \ge 1.5\). Comparing this result with the provided options, \(x = 14\) matches the fourth option. Therefore, the correct solution is \(\boxed{14}\).
Square root equations like this one are foundational for understanding constraints in algebra. In engineering, for example, if the velocity \(v\) of an object is related to kinetic energy \(K\) by \(v = \sqrt{2K/m}\), solving for the mass or energy requires the same process of squaring both sides to isolate the variable. Mastering this technique allows you to manipulate formulas and solve for unknowns in various physical and mathematical models.