Let's analyze Joana's conclusion based on her steps for solving the quadratic equation \(4x^2 - 4x - 3 = 0\).
Step 1: Joana rewrote the middle term, splitting \(-4x\) into \(2x\) and \(-6x\), which is correct since \(2x - 6x = -4x\). The equation now reads:
\[ 4x^2 + 2x - 6x - 3 = 0 \]
Step 2: She grouped the terms:
\[ (4x^2 + 2x) + (-6x - 3) = 0 \]
Step 3: She factored each grouping:
\[ x(4x + 2) - 3(2x + 1) = 0 \]
At this point, Joana noticed that the two groupings \(x(4x + 2)\) and \(-3(2x + 1)\) do not share a common binomial factor.
Joana's conclusion that the quadratic equation has no solutions is not correct. The lack of a common factor in the groupings does not imply that there are no solutions to the quadratic equation.
Quadratic equations can still have solutions even if they don't factor neatly. The equation can also be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For \(4x^2 - 4x - 3 = 0\), we can take \(a = 4\), \(b = -4\), and \(c = -3\):
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-3)}}{2(4)} \] \[ x = \frac{4 \pm \sqrt{16 + 48}}{8} \] \[ x = \frac{4 \pm \sqrt{64}}{8} \] \[ x = \frac{4 \pm 8}{8} \]
From this, we find two possible solutions:
- \(x = \frac{12}{8} = \frac{3}{2}\)
- \(x = \frac{-4}{8} = -\frac{1}{2}\)
So, the correct response about Joana's conclusion is:
No, because she can continue the steps even if the two groupings have no shared binomial.