To determine the solutions to the system of equations given by \( y = -x^2 \) and \( y = 12x - 3 \), we will set the two equations equal to each other and solve for \( x \):
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Set the equations equal: \[ -x^2 = 12x - 3 \]
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Rearrange to form a standard quadratic equation: \[ -x^2 - 12x + 3 = 0 \] or multiplying through by -1: \[ x^2 + 12x - 3 = 0 \]
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Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1 \), \( b = 12 \), and \( c = -3 \): \[ b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot (-3) = 144 + 12 = 156 \]
Thus, we can solve for \( x \): \[ x = \frac{-12 \pm \sqrt{156}}{2} \] Simplifying \( \sqrt{156} \): \[ \sqrt{156} = \sqrt{4 \cdot 39} = 2\sqrt{39} \]
Therefore, \[ x = \frac{-12 \pm 2\sqrt{39}}{2} = -6 \pm \sqrt{39} \]
This gives us two \( x \) values: \[ x_1 = -6 + \sqrt{39}, \quad x_2 = -6 - \sqrt{39} \]
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To find the corresponding \( y \) values, we can substitute these \( x \) values back into either of the original equations. We’ll use \( y = 12x - 3 \): For \( x_1 \): \[ y_1 = 12(-6 + \sqrt{39}) - 3 = -72 + 12\sqrt{39} - 3 = -75 + 12\sqrt{39} \] For \( x_2 \): \[ y_2 = 12(-6 - \sqrt{39}) - 3 = -72 - 12\sqrt{39} - 3 = -75 - 12\sqrt{39} \]
Therefore, the solutions to the system are: \[ \left(-6 + \sqrt{39}, -75 + 12\sqrt{39}\right) \quad \text{and} \quad \left(-6 - \sqrt{39}, -75 - 12\sqrt{39}\right) \]
None of the provided options exactly match those calculated solutions. If you provide further context or clarify the options, I could assist more accurately!