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Solve the quadratic equation using the quadratic formula. Find only real solutions. If there are no real solutions, enter "DNE" in both boxes. \( 1+x-x^{2}=0 \)

Ask by Boyd Burns. in the United States
Jan 23,2025

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Answer

The solutions are \( x = \frac{1 + \sqrt{5}}{2} \) and \( x = \frac{1 - \sqrt{5}}{2} \).

Solution

To solve the quadratic equation \( 1 + x - x^{2} = 0 \) using the quadratic formula, follow these steps: 1. **Rewrite the Equation in Standard Form:** \[ -x^{2} + x + 1 = 0 \quad \text{or} \quad x^{2} - x - 1 = 0 \] Here, \( a = 1 \), \( b = -1 \), and \( c = -1 \). 2. **Apply the Quadratic Formula:** \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Substitute the values of \( a \), \( b \), and \( c \): \[ x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \] 3. **Determine the Real Solutions:** \[ x = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{5}}{2} \] Both solutions are real numbers since the discriminant (\( \sqrt{5} \)) is positive. **Final Answer:** \[ x = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{5}}{2} \]

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Beyond the Answer

To solve the quadratic equation \( 1 + x - x^{2} = 0 \), we first rewrite it in standard form: \( -x^{2} + x + 1 = 0 \), or multiplying through by -1, \( x^{2} - x - 1 = 0 \). Now, we can apply the quadratic formula, which is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -1 \), and \( c = -1 \). Calculating the discriminant: \( b^2 - 4ac = (-1)^{2} - 4(1)(-1) = 1 + 4 = 5 \). Now, plug values into the quadratic formula: \[ x = \frac{-(-1) \pm \sqrt{5}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}. \] Thus, the real solutions are: \[ x = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{5}}{2}. \] Final answers: \( x_1 = \frac{1 + \sqrt{5}}{2} \), \( x_2 = \frac{1 - \sqrt{5}}{2} \).

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