Solve sqrt(14-x) = sqrt(x-4) + sqrt(x-1)

For the square roots to be defined, we must have:
$14-x \ge 0 \implies x \le 14$
$x-4 \ge 0 \implies x \ge 4$
$x-1 \ge 0 \implies x \ge 1$
Combining these conditions, the domain is $4 \le x \le 14$.

For this equation to hold, we must have $19-3x \ge 0$, which means $3x \le 19$, or $x \le \frac{19}{3} \approx 6.33$.

The two potential solutions are:

For $x_1 = 13.8$: This value satisfies $4 \le x \le 14$, but it does not satisfy $x \le \frac{19}{3}$ (since $13.8 > 6.33$). Thus, $x=13.8$ is an extraneous solution.
For $x_2 = 5$: This value satisfies $4 \le x \le 14$ and $x \le \frac{19}{3}$.
Substitute $x=5$ into the original equation:
$\sqrt{14-5} = \sqrt{5-4} + \sqrt{5-1}$
$\sqrt{9} = \sqrt{1} + \sqrt{4}$
$3 = 1 + 2$
$3 = 3$
This is true.

The solution is $\boxed{x=5}$.

For the square roots to be defined, we must have:
$4x+1 \ge 0 \implies 4x \ge -1 \implies x \ge -\frac{1}{4}$
$2x-3 \ge 0 \implies 2x \ge 3 \implies x \ge \frac{3}{2}$
Combining these conditions, the domain is $x \ge \frac{3}{2}$.

Divide by 2:

For this equation to hold, we must have $9-3x \ge 0$, which means $3x \le 9$, or $x \le 3$.

We need two numbers that multiply to 84 and add to -44. These numbers are -2 and -42.

The two potential solutions are:

For $x_1 = 2$: This value satisfies $x \ge \frac{3}{2}$ and $x \le 3$.
Substitute $x=2$ into the original equation:
$\sqrt{4(2)+1} + \sqrt{2(2)-3} = 4$
$\sqrt{8+1} + \sqrt{4-3} = 4$
$\sqrt{9} + \sqrt{1} = 4$
$3 + 1 = 4$
$4 = 4$
This is true.
For $x_2 = 42$: This value satisfies $x \ge \frac{3}{2}$, but it does not satisfy $x \le 3$ (since $42 > 3$). Thus, $x=42$ is an extraneous solution.

The solution is $\boxed{x=