To solve the quadratic equation \(6x^2 - 3x + 6 = 0\), we can use the quadratic formula, which is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \(a = 6\), \(b = -3\), and \(c = 6\).
- First, calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = (-3)^2 = 9 \] \[ 4ac = 4 \cdot 6 \cdot 6 = 144 \] \[ b^2 - 4ac = 9 - 144 = -135 \]
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Since the discriminant is negative, the solutions will be complex numbers.
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Now, substitute \(a\), \(b\), and the discriminant into the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{-135}}{2 \cdot 6} \] \[ x = \frac{3 \pm \sqrt{-135}}{12} \] \[ x = \frac{3 \pm i\sqrt{135}}{12} \]
- We can simplify \(\sqrt{135}\):
\[ \sqrt{135} = \sqrt{9 \times 15} = 3\sqrt{15} \]
- Substituting this back in gives:
\[ x = \frac{3 \pm 3i\sqrt{15}}{12} \] \[ x = \frac{1 \pm i\sqrt{15}}{4} \]
- Therefore, the solutions can be expressed as:
\[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4} i \]
Based on the answer choices you provided, 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
1 answer