Joana solved the equation 4x2−4x−3=0. 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 4x2−4x−3=0 has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.

Question

Step 1: 4x2+2x+(−6x)−3=0

Step 2: (4x2+2x)+(−6x−3)=0

Step 3: x(4x+2)+(−3(2x+1))=0

(1 point)
Responses

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

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 negative 12 that add up to negative 4 .

Answer 1

To evaluate Joana's conclusion, let's explore her work step by step.

1. Original Equation: \(4x^2 - 4x - 3 = 0\)

2. Step 1: Joana rewrote the equation as \(4x^2 + 2x - 6x - 3 = 0\). This splitting of the middle term is valid because \(2x - 6x = -4x\).

3. Step 2: She grouped the terms as \((4x^2 + 2x) + (-6x - 3) = 0\).

4. Step 3: She factored out the common factors:

   • From the first grouping \((4x^2 + 2x)\), she factored out \(2x\), giving \(2x(2x + 1)\).
   • From the second grouping \((-6x - 3)\), she actually factored incorrectly and wrote it as \(-3(2x + 1)\), which is correct in terms of factoring but doesn't lead to a common factor.

At this point, it looks like she didn't notice that both groups could still produce valid results despite not sharing a common binomial due to her earlier splitting step.

Next, let’s summarize the important points:

• If she had factored correctly in both groupings, she would have obtained: \[ 2x(2x + 1) - 3(2x + 1) = 0 \]
• This could be factored to: \[ (2x - 3)(2x + 1) = 0 \]
• Now we can solve for \(x\): \[ 2x - 3 = 0 \Rightarrow x = \frac{3}{2} \] \[ 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \]

Since there are indeed solutions to the equation \(4x^2 - 4x - 3 = 0\), Joana's conclusion that the quadratic equation has no solutions is incorrect.

Based on the options provided, the correct response is:

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