To solve the equation \(-\sqrt{7x + 4} = \sqrt{1 - x} + 3\), we start by isolating the square roots on one side:
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Move all non-square root terms to one side: \[ -\sqrt{7x + 4} - 3 = \sqrt{1 - x} \]
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Next, we square both sides to eliminate the square roots. Remember: when squaring both sides, we need to be careful about extraneous solutions: \[ (-\sqrt{7x + 4} - 3)^2 = (\sqrt{1 - x})^2 \]
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Expanding both sides gives us: \[ (7x + 4) + 6\sqrt{7x + 4} + 9 = 1 - x \] \[ 7x + 13 + 6\sqrt{7x + 4} = 1 - x \]
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Rearranging it gives: \[ 6\sqrt{7x + 4} = 1 - x - 7x - 13 \] \[ 6\sqrt{7x + 4} = -8x - 12 \]
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Dividing both sides by 6: \[ \sqrt{7x + 4} = \frac{-8x - 12}{6} \] Simplifying the right side: \[ \sqrt{7x + 4} = -\frac{4x + 6}{3} \]
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Now, we square again to eliminate the square root: \[ 7x + 4 = \left(-\frac{4x + 6}{3}\right)^2 \] \[ 7x + 4 = \frac{(4x + 6)^2}{9} \]
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Multiplying through by 9 to clear the fraction: \[ 9(7x + 4) = (4x + 6)^2 \] \[ 63x + 36 = 16x^2 + 48x + 36 \]
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Simplifying gives: \[ 0 = 16x^2 - 15x \] Factoring out \(x\): \[ 0 = x(16x - 15) \]
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Setting each factor to zero gives: \[ x = 0 \quad \text{or} \quad 16x - 15 = 0 \implies x = \frac{15}{16} \]
Now we must check for extraneous solutions by substituting back into the original equation.
For \(x = 0\): \[ -\sqrt{7(0) + 4} = \sqrt{1 - 0} + 3 \] \[ -2 = 1 + 3 \implies -2 \neq 4 \quad (\text{not a solution}) \]
For \(x = \frac{15}{16}\): \[ -\sqrt{7\left(\frac{15}{16}\right) + 4} = \sqrt{1 - \frac{15}{16}} + 3 \] Calculating: \[ 7\left(\frac{15}{16}\right) + 4 = \frac{105}{16} + \frac{64}{16} = \frac{169}{16} \implies -\frac{13}{4} \] \[ \frac{1}{4} + 3 = \frac{1}{4} + \frac{12}{4} = \frac{13}{4} \] So, \[ -\frac{13}{4} \neq \frac{13}{4} \quad (\text{not a solution}) \]
Since neither solution is valid in the original equation, the equation has no solution.
Thus the final answer is: \[ \boxed{\text{No solution}} \]