To determine whether Joana's conclusion is correct, let's analyze the steps she took to solve the quadratic equation \(4x^2 - 4x - 3 = 0\).
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Step 1: Joana rewrote the equation as \(4x^2 + 2x - 6x - 3 = 0\). This step is valid as she is preparing to factor by grouping.
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Step 2: She grouped the first two terms and the last two terms as \((4x^2 + 2x) + (-6x - 3) = 0\). This step is also valid.
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Step 3: When she factored each grouping, she factored out \(2x\) from the first grouping, leading to:
- First group: \(2x(2x + 1)\)
- Second group: \(-3(2x + 1)\)
This results in the expression: \[ 2x(2x + 1) - 3(2x + 1) = 0 \]
At this step, she notices that the two groupings do share a common binomial, which is \((2x + 1)\). Therefore, she can factor this as: \[ (2x + 1)(2x - 3) = 0 \]
Now, to find the solutions to the original equation, we can set each factor to zero:
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\(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)
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\(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\)
Thus, the quadratic equation actually has two solutions: \(x = -\frac{1}{2}\) and \(x = \frac{3}{2}\).
Based on the analysis:
- Joana's conclusion that the equation has no solutions is not correct.
Therefore, the answer to the question is:
No, because she can still continue the steps even if the two groupings have no shared binomial.