Joana solved the equation 4x2−4x−3=0

Question

. She followed the steps to solve a quadratic equation by grouping. However, in Step 3 she noticed that the two groupings did not share a common binomial, as shown in her work below. She concluded that the quadratic equation has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.

Step 1: 4x2+2x+(−6x)−3=0

Step 2: (4x2+2x)+(−6x−3)=0

Step 3: x(4x+2x)+(−3(2x+1))=0


(1 point)
Responses

No, because she did not factor out the GCF from the first grouping.
No, because she did not factor out the GCF from the first grouping.

No, because she can still continue the steps even if the two groupings have no shared
binomial.
No, because she can still continue the steps even if the two groupings have no shared binomial.

Yes, because there are no factors of −12 that add up to −4.
Yes, because there are no factors of −12 that add up to −4.

Yes, because the two groupings do not share a common binomial.

Answer 1

To analyze Joana's conclusion that the quadratic equation \(4x^2 - 4x - 3 = 0\) has no solutions based on her steps, let's examine her work.

1. Step 1: Grouping
   Joana rewrites the equation as \(4x^2 + 2x - 6x - 3 = 0\). Here, she correctly breaks down the middle term \(-4x\) into \(+2x\) and \(-6x\).

2. Step 2: Grouping Terms
   She groups the terms as \((4x^2 + 2x) + (-6x - 3) = 0\). This grouping is valid.

3. Step 3: Factor Out GCF
   Joana's attempt to factor gives:

   • From the first grouping \(4x^2 + 2x\), she factored out \(2x\) but stated \(x(4x + 2x)\), which is incorrect. The correct factorization should be \(2x(2x + 1)\).
   • From the second grouping \(-6x - 3\), she factored out \(-3\) correctly to get \(-3(2x + 1)\).

Her groupings resulted in: \[ 2x(2x + 1) - 3(2x + 1) = 0 \]

At this point, we notice that there is a common factor \((2x + 1)\), which Joana mistakenly claims does not exist. Therefore, she should correctly factor it out:

\[ (2x + 1)(2x - 3) = 0 \]

After this correct factoring, we set each factor to zero to find the solutions:

1. \(2x + 1 = 0 \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}\)
2. \(2x - 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}\)

Thus, the equation has two solutions.

In conclusion, Joana's assertion that the quadratic has no solutions is incorrect. The correct answer is:

No, because she can still continue the steps even if the two groupings have no shared binomial.