Solve: $\sqrt{2a-5} = a-4$
To solve the radical equation \(\sqrt{2a - 5} = a - 4\), we must first identify the domain of validity. The square root function \(\sqrt{x}\) is defined only for non-negative real numbers, so we must have \(2a - 5 \geq 0\), which implies \(a \geq 2.5\). Additionally, since \(\sqrt{2a - 5}\) is always non-negative, the right-hand side of the equation must also be non-negative, giving \(a - 4 \geq 0\), or \(a \geq 4\). Combining these conditions, any valid solution for \(a\) must satisfy \(a \geq 4\).
Next, we eliminate the square root by squaring both sides of the equation:
\[(\sqrt{2a - 5})^2 = (a - 4)^2\]
Expanding the right side gives:
\[2a - 5 = a^2 - 8a + 16\]
Rearranging this into standard quadratic form \(ax^2 + bx + c = 0\):
\[a^2 - 10a + 21 = 0\]
We now solve the quadratic equation \(a^2 - 10a + 21 = 0\) by factoring. Looking for two numbers that multiply to \(21\) and add to \(-10\), we find \(-7\) and \(-3\):
\[(a - 7)(a - 3) = 0\]
This gives two potential solutions for \(a\):
\[a = 7 \quad \text{or} \quad a = 3\]
Finally, we verify our potential solutions against the constraint \(a \geq 4\) established in step 1.
Checking \(a = 7\) in the original equation:
LHS: \(\sqrt{2(7) - 5} = \sqrt{14 - 5} = \sqrt{9} = 3\)
RHS: \(7 - 4 = 3\)
The equation holds. Thus, the only valid solution is \(\boxed{a = 7}\).
Radical equations often appear in physics and engineering when calculating distances or determining the intersection of pathways. For instance, if a projectile's speed is defined as a function of time involving a square root, setting this equal to a time-dependent expression allows you to find the exact moment the projectile reaches a specific speed. This systematic method of squaring, solving, and checking for extraneous solutions ensures accurate results in real-world modeling.