To find which of the options is a solution to the equation \( x + x^2 - 1 = 0 \), we can substitute each option into the equation and check if it satisfies the equation.
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Option (A): \( x = -1 \) \[ -1 + (-1)^2 - 1 = -1 + 1 - 1 = -1 \quad (\text{not } 0) \]
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Option (B): \( x = 0 \) \[ 0 + 0^2 - 1 = 0 + 0 - 1 = -1 \quad (\text{not } 0) \]
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Option (C): \( x = \frac{1}{2} \) \[ \frac{1}{2} + \left(\frac{1}{2}\right)^2 - 1 = \frac{1}{2} + \frac{1}{4} - 1 \] Converting to a common denominator (4): \[ \frac{2}{4} + \frac{1}{4} - \frac{4}{4} = \frac{2 + 1 - 4}{4} = \frac{-1}{4} \quad (\text{not } 0) \]
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Option (D): \( x = 1 \) \[ 1 + 1^2 - 1 = 1 + 1 - 1 = 1 \quad (\text{not } 0) \]
None of the options seem to satisfy the equation \( x + x^2 - 1 = 0 \).
To double-check, let's actually solve the equation \( x + x^2 - 1 = 0 \): Rearranging gives us: \[ x^2 + x - 1 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1, b = 1, c = -1 \): \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
Since the roots are: \[ x = \frac{-1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{5}}{2} \]
Neither of these roots matches any of the options provided. Therefore, none of the options (A, B, C, D) is a solution to the equation \( x + x^2 - 1 = 0 \).
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