To solve the quadratic equation \(6x^2 - 3x + 6 = 0\), we will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 6\), \(b = -3\), and \(c = 6\).
- Calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 = (-3)^2 = 9 \] \[ 4ac = 4 \cdot 6 \cdot 6 = 144 \] \[ b^2 - 4ac = 9 - 144 = -135 \]
Since the discriminant is negative, the solutions will include imaginary numbers.
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Substitute the values into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{-135}}{2 \cdot 6} \] \[ x = \frac{3 \pm \sqrt{-135}}{12} \]
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Simplify \(\sqrt{-135}\): \[ \sqrt{-135} = i\sqrt{135} = i\sqrt{9 \cdot 15} = 3i\sqrt{15} \]
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Now substitute back: \[ x = \frac{3 \pm 3i\sqrt{15}}{12} \] \[ = \frac{3}{12} \pm \frac{3i\sqrt{15}}{12} \] \[ = \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
Thus, the solutions can be expressed in the form \(a \pm bi\): \[ \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
The correct response is: Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i
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