To solve the equation \( 2x - 1 = \sqrt{8 - x} \), we first square both sides to eliminate the square root.
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Start with the equation: \[ 2x - 1 = \sqrt{8 - x} \]
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Square both sides: \[ (2x - 1)^2 = (8 - x) \]
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Expand the left side: \[ 4x^2 - 4x + 1 = 8 - x \]
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Rearrange the equation: \[ 4x^2 - 4x + x + 1 - 8 = 0 \] \[ 4x^2 - 3x - 7 = 0 \]
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Now, we can solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4 \), \( b = -3 \), and \( c = -7 \).
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Calculate the discriminant: \[ b^2 - 4ac = (-3)^2 - 4(4)(-7) = 9 + 112 = 121 \]
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Now substitute back into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{121}}{2(4)} = \frac{3 \pm 11}{8} \]
This gives us two solutions: \[ x = \frac{14}{8} = \frac{7}{4} \quad \text{and} \quad x = \frac{-8}{8} = -1 \]
So the potential solutions are \( x = \frac{7}{4} \) and \( x = -1 \).
Next, we need to check for extraneous solutions by substituting these values back into the original equation.
Checking the solutions:
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Checking \( x = \frac{7}{4} \): \[ \text{Left side: } 2\left(\frac{7}{4}\right) - 1 = \frac{14}{4} - 1 = \frac{14}{4} - \frac{4}{4} = \frac{10}{4} = \frac{5}{2} \] \[ \text{Right side: } \sqrt{8 - \frac{7}{4}} = \sqrt{\frac{32}{4} - \frac{7}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} \] Since both sides are equal, \( x = \frac{7}{4} \) is a valid solution.
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Checking \( x = -1 \): \[ \text{Left side: } 2(-1) - 1 = -2 - 1 = -3 \] \[ \text{Right side: } \sqrt{8 - (-1)} = \sqrt{8 + 1} = \sqrt{9} = 3 \] Since the left side is \(-3\) and the right side is \(3\), they are not equal. Thus, \( x = -1 \) is an extraneous solution.
Conclusion:
The only valid solution is \( x = \frac{7}{4} \). Among the options provided:
Answer: A \( x = \frac{7}{4} \).
1 answer