Solve the radical equations: 22. sqrt(2 − x) = x − 2 23. sqrt(43 − x) − x + 1 = 0

For the square root $\sqrt{2-x}$ to be defined, $2-x \ge 0 \implies x \le 2$.
For the right side $x-2$ to be non-negative (as it equals a square root), $x-2 \ge 0 \implies x \ge 2$.
Combining these, the only possible value for $x$ is $x=2$.

Substitute $x=2$ into the equation:

This is true, so $x=2$ is a valid solution.

Alternatively, by squaring both sides:

This gives two potential solutions: $x = 1$ or $x = 2$.

For $x=1$:

This is false, so $x=1$ is an extraneous solution.

For $x=2$:

This is true, so $x=2$ is a valid solution.

The final answer is $\boxed{x = 2}$.

23. Solve $\sqrt{43 - x} - x + 1 = 0$

For the square root $\sqrt{43-x}$ to be defined, $43-x \ge 0 \implies x \le 43$.
For the right side $x-1$ to be non-negative, $x-1 \ge 0 \implies x \ge 1$.
So, any valid solution must satisfy $1 \le x \le 43$.

We look for two numbers that multiply to -42 and add to -1. These are -7 and 6.

This gives two potential solutions: $x = 7$ or $x = -6$.

For $x=7$:
This satisfies $1 \le 7 \le 43$.
Substitute into the original equation:

This is true, so $x=7$ is a valid solution.

For $x=-6$:
This does not satisfy $x \ge 1$ (since $-6 < 1$). So $x=-6$ is an extraneous solution.
Alternatively, substitute into the original equation:

This is false, so $x=-6$ is an extraneous solution.

The final answer is $\boxed{x = 7}$.

24. Solve $2\sqrt{x + 6} + 2 = x$

For the square root $\sqrt{x+6}$ to be defined, $x+6 \ge 0 \implies x \ge -6$.
For the right side $x-2$ to be non-negative (as it equals twice a square root), $x-2 \ge 0 \implies x \ge 2$.
So, any valid solution must satisfy $x \ge 2$.

We look for two numbers that multiply to -20 and add to -8. These are -10 and 2.

This gives two potential solutions: $x = 10$ or $x = -2$.

For $x=10$:
This satisfies $x \ge 2$ (since $10 \ge 2$).
Substitute into the original equation:

This is true, so $x=10$ is a valid solution.

For $x=-2$:
This does not satisfy $x \ge 2$ (since $-2 < 2$). So $x=-2$ is an extraneous solution.
Alternatively, substitute into the original equation:

This is false, so $x=-2$ is an extraneous solution.

The final answer is $\boxed{x = 10}$.