To solve the radical equation
\[ -\sqrt{-7x + 4} = \sqrt{1 - x} + 3, \]
we'll first isolate the square root on one side. Let's rewrite the equation:
\[ -\sqrt{-7x + 4} - 3 = \sqrt{1 - x}. \]
Now, we square both sides to eliminate the square roots. However, be careful with the negative sign. To ensure we square correctly, note that squaring both sides will involve negatives, so let's switch the terms:
\[ \sqrt{-7x + 4} = -(\sqrt{1 - x} + 3). \]
Squaring both sides results in:
\[ -7x + 4 = (-(\sqrt{1 - x} + 3))^2. \]
Expanding the right-hand side:
\[ -7x + 4 = (-(\sqrt{1 - x}) - 3)^2 = (\sqrt{1 - x} + 3)^2 = (1 - x) + 6\sqrt{1 - x} + 9 = 10 - x + 6\sqrt{1 - x}. \]
Now we have:
\[ -7x + 4 = 10 - x + 6\sqrt{1 - x}. \]
Rearranging this gives:
\[ -7x + x + 4 - 10 = 6\sqrt{1 - x}. \]
Simplifying the left side:
\[ -6x - 6 = 6\sqrt{1 - x}. \]
Now divide both sides by 6:
\[ -x - 1 = \sqrt{1 - x}. \]
Next, square both sides again:
\[ (-x - 1)^2 = 1 - x. \]
Expanding the left side:
\[ x^2 + 2x + 1 = 1 - x. \]
Now rearranging gives:
\[ x^2 + 2x + x + 1 - 1 = 0, \]
which simplifies to
\[ x^2 + 3x = 0. \]
Factoring gives:
\[ x(x + 3) = 0. \]
Thus, the possible solutions are:
\[ x = 0 \quad \text{or} \quad x = -3. \]
Now we need to check both solutions in the original equation:
- For \(x = 0\):
\[ -\sqrt{-7(0) + 4} = 1 - 0 + 3, \] \[ -\sqrt{4} = 4, \] \[ -2 \neq 4 \quad \text{(not a solution)}. \]
- For \(x = -3\):
\[ -\sqrt{-7(-3) + 4} = 1 - (-3) + 3, \] \[ -\sqrt{21 + 4} = 1 + 3, \] \[ -\sqrt{25} = 4, \] \[ -5 = 4 \quad \text{(not a solution)}. \]
Since both possible solutions are not valid, we end up with no solution:
The final answer is that there are no valid solutions for the given equation.