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 4x2−4x−3=0
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+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.

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 .

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.

Answer 1

To determine whether Joana's conclusion is correct, let's analyze the steps she took to solve the quadratic equation \(4x^2 - 4x - 3 = 0\).

1. 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.

2. 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.

3. 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:

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

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

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.