To solve the equation \( \sqrt{x-1} + 4 = x - 3 \), follow these steps:
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Isolate the square root: \[ \sqrt{x-1} = x - 3 - 4 \] \[ \sqrt{x-1} = x - 7 \]
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Square both sides to eliminate the square root: \[ (\sqrt{x-1})^2 = (x - 7)^2 \] \[ x - 1 = (x - 7)(x - 7) \] \[ x - 1 = x^2 - 14x + 49 \]
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Rearrange the equation: \[ 0 = x^2 - 14x + 49 - x + 1 \] \[ 0 = x^2 - 15x + 50 \]
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Factor or use the quadratic formula to solve for \( x \): The quadratic is \( x^2 - 15x + 50 = 0 \). We can factor this as: \[ (x - 10)(x - 5) = 0 \] which gives us: \[ x = 10 \quad \text{or} \quad x = 5 \]
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Check for extraneous solutions:
For \( x = 10 \): \[ \sqrt{10-1} + 4 = 3 + 4 = 7 \] \[ 10 - 3 = 7 \] Thus, \( x = 10 \) is a valid solution.
For \( x = 5 \): \[ \sqrt{5-1} + 4 = 2 + 4 = 6 \] \[ 5 - 3 = 2 \] Thus, \( x = 5 \) does not satisfy the original equation.
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Final solution: The only solution to the equation is \[ \boxed{10} \]
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