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 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.

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 can still continue the steps even if the two groupings have no shared binomial.

Answer 1

Joana's conclusion is not correct. The fact that her grouping led to no common binomial does not mean that the quadratic equation has no solutions. Let's analyze her steps:

1. In Step 1, Joana rewrote the equation as \(4x^2 + 2x - 6x - 3 = 0\).
2. In Step 2, she grouped the terms: \((4x^2 + 2x) + (-6x - 3) = 0\).
3. In Step 3, Joana factored the first group as \(2x(2x + 1)\) and the second group incorrectly—one possible correction would have been to factor out \(-3\) to get \(-3(2x + 1)\).

Even though Joana encountered issues with grouping, quadratic equations can also be solved using the quadratic formula when factoring is complex or unhelpful. The equation \(4x^2 - 4x - 3 = 0\) should have solutions, which can be confirmed by using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 4\), \(b = -4\), and \(c = -3\):

\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \] \[ x = \frac{4 \pm \sqrt{16 + 48}}{8} \] \[ x = \frac{4 \pm \sqrt{64}}{8} \] \[ x = \frac{4 \pm 8}{8} \]

This gives two solutions:

1. \(x = \frac{12}{8} = 1.5\)
2. \(x = \frac{-4}{8} = -0.5\)

Thus, the answers exist.

Therefore, the correct response to Joana's conclusion is:

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