To solve the equation \(3(2x-1)^{1/2} (4x-2) + 2 = 0\), let's first simplify and isolate the expression step by step.
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Start by isolating the square root term. Subtract 2 from both sides: \[ 3(2x-1)^{1/2} (4x-2) = -2 \]
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Now we notice that \(4x - 2 = 2(2x - 1)\). So we can rewrite the equation: \[ 3(2x-1)^{1/2} \cdot 2(2x - 1) = -2 \]
This simplifies to: \[ 6(2x-1)^{3/2} = -2 \]
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Now, divide both sides by 6: \[ (2x-1)^{3/2} = -\frac{1}{3} \]
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The left side, \((2x-1)^{3/2}\), is non-negative for all \(x\) where \(2x-1\) is defined, which means \((2x-1)^{3/2} \geq 0\). However, the right side is negative \(-\frac{1}{3}\).
Since a non-negative number cannot equal a negative number, there are no solutions to the equation.
Thus, the solution to the equation is: \[ \text{No solution} \]