Joana solved the equation 4x^2−4x−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: 4x^2+2x+(−6x)−3=0

Step 2:(4x^2+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 determine whether Joana's conclusion is correct, let’s first analyze her steps:

1. Rewrite the equation: Joana rewrote \(4x^2 - 4x - 3 = 0\) as \(4x^2 + 2x - 6x - 3 = 0\), which is valid.

2. Group the equation: She grouped it as \((4x^2 + 2x) + (-6x - 3) = 0\). This is also valid grouping.

3. Factor the groups: In Step 3, she tries to factor out the greatest common factor (GCF) from each grouping. However, her factorizations are incorrect:

   • For \(4x^2 + 2x\), the correct factorization should be \(2x(2x + 1)\), not \(x(4x + 2x)\).
   • For \(-6x - 3\), the correct factorization is \(-3(2x + 1)\), but it appears she did not notice that the grouped terms \(2x + 1\) are actually the same.

When factoring the groups correctly: \[ 2x(2x + 1) - 3(2x + 1) = 0 \] This indeed shares the common binomial \(2x + 1\).

4. Set the common factor to 0: This leads to: \[ (2x - 3)(2x + 1) = 0 \] Thus, we can find the solutions for \(x\):

• From \(2x - 3 = 0\), we find \(x = \frac{3}{2}\).
• From \(2x + 1 = 0\), we find \(x = -\frac{1}{2}\).

From this analysis, we conclude:

• Joana's initial conclusion that there are no solutions is incorrect, as the quadratic equation actually has two solutions.

So, the correct response is: No, because she can still continue the steps even if the two groupings have no shared binomial.