To solve the equation \( \sqrt{x} = x - 2 \), we will first isolate the square root and then square both sides to eliminate the square root.
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Isolate the square root and square both sides: \[ \sqrt{x} = x - 2 \] Square both sides: \[ (\sqrt{x})^2 = (x - 2)^2 \] This gives us: \[ x = (x - 2)(x - 2) \] Expanding the right-hand side: \[ x = x^2 - 4x + 4 \]
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Rearranging the equation: Move all terms to one side: \[ 0 = x^2 - 4x + 4 - x \] Simplify: \[ 0 = x^2 - 5x + 4 \] This is a quadratic equation in the standard form.
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Factor the quadratic: Rewrite it as: \[ 0 = (x - 1)(x - 4) \]
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Setting each factor to zero gives the solutions: \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
Now we have two potential solutions: \( x = 1 \) and \( x = 4 \).
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Check for extraneous solutions: We need to check each potential solution in the original equation \( \sqrt{x} = x - 2 \).
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For \( x = 1 \): \[ \sqrt{1} = 1 - 2 \quad \Rightarrow \quad 1 = -1 \quad \text{(false)} \] So, \( x = 1 \) is an extraneous solution.
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For \( x = 4 \): \[ \sqrt{4} = 4 - 2 \quad \Rightarrow \quad 2 = 2 \quad \text{(true)} \] So, \( x = 4 \) is a valid solution.
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Conclusion: The valid solution to the equation \( \sqrt{x} = x - 2 \) is \( \boxed{4} \). The solution \( x = 1 \) is an extraneous solution.