To solve the equation \( 3(2x - 1) \frac{1}{2} (4x - 2) + 2 = 0 \), let's first clarify the format of the equation.
It seems there might be some confusion regarding the placement of the \( \frac{1}{2} \). I will interpret the expression as:
\[ 3(2x - 1) \cdot \frac{1}{2}(4x - 2) + 2 = 0 \]
Now let's simplify it step-by-step.
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Distribute the terms inside \( \frac{1}{2}(4x - 2) \):
\[ \frac{1}{2}(4x - 2) = \frac{4x}{2} - \frac{2}{2} = 2x - 1 \]
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Now substitute this back into the equation:
\[ 3(2x - 1)(2x - 1) + 2 = 0 \]
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Expand \( 3(2x - 1)(2x - 1) \):
\[ (2x - 1)(2x - 1) = (2x - 1)^2 = 4x^2 - 4x + 1 \]
Therefore,
\[ 3(4x^2 - 4x + 1) = 12x^2 - 12x + 3 \]
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Substituting this back gives:
\[ 12x^2 - 12x + 3 + 2 = 0 \]
Which simplifies to:
\[ 12x^2 - 12x + 5 = 0 \]
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Now, apply the quadratic formula:
The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 12\), \(b = -12\), and \(c = 5\).
- Calculate the discriminant:
\[ b^2 - 4ac = (-12)^2 - 4 \cdot 12 \cdot 5 = 144 - 240 = -96 \]
Since the discriminant is negative (\(-96\)), there are no real solutions.
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Thus, the final answer is:
The equation \( 3(2x - 1) \cdot \frac{1}{2}(4x - 2) + 2 = 0 \) has no real solutions. The solutions are complex numbers:
\[ x = \frac{12 \pm \sqrt{-96}}{24} = \frac{12 \pm 4i\sqrt{6}}{24} = \frac{1 \pm \frac{i\sqrt{6}}{6}}{2} = \frac{1}{2} \pm \frac{i\sqrt{6}}{12} \]
So the complex solutions can be expressed as:
\[ x = \frac{1}{2} + \frac{i\sqrt{6}}{12}, \quad \frac{1}{2} - \frac{i\sqrt{6}}{12} \]
That's your answer!