Joana solved the equation 4x2−4x−3=0

Question

4
𝑥
2
−
4
𝑥
−
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
4
𝑥
2
−
4
𝑥
−
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
4
𝑥
2
+
2
𝑥
+
(
−
6
𝑥
)
−
3
=
0

Step 2: (4x2+2x)+(−6x−3)=0
(
4
𝑥
2
+
2
𝑥
)
+
(
−
6
𝑥
−
3
)
=
0

Step 3: x(4x+2)+(−3(2x+1))=0
𝑥
(
4
𝑥
+
2
)
+
(
−
3
(
2
𝑥
+
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.

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

Answer 1

Joana's conclusion is not correct. Let's analyze her work step by step and clarify why.

Review of Joana's Work

1. Step 1: Joana rewrites the equation \(4x^2 - 4x - 3 = 0\) as \(4x^2 + 2x - 6x - 3 = 0\). This is a valid transformation.

2. Step 2: She groups the terms as \((4x^2 + 2x) + (-6x - 3) = 0\). This is also correct.

3. Step 3: Joana attempts to factor out the greatest common factors (GCF) from each grouping, resulting in \(x(4x + 2) + (-3(2x + 1)) = 0\). However, the two groupings, \( (4x + 2) \) and \( (-3(2x + 1)) \), do not share a common factor of the same binomial, which leads her to a conclusion of no solutions.

Explanation of the Conclusion

The critical mistake in her reasoning is that not sharing a common binomial does not imply that the original quadratic equation has no solutions.

Finding the Solutions

Instead of concluding that there are no solutions, let's find the solutions correctly either by factoring or using the quadratic formula:

1. Quadratic Formula: The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Here, \(a = 4\), \(b = -4\), \(c = -3\).

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

2. Calculating the Discriminant: \[ b^2 - 4ac = (-4)^2 - 4(4)(-3) = 16 + 48 = 64 \]

3. Calculating the Roots: \[ x = \frac{-(-4) \pm \sqrt{64}}{2(4)} = \frac{4 \pm 8}{8} \]

   This gives us: \[ x = \frac{12}{8} = \frac{3}{2} \quad \text{and} \quad x = \frac{-4}{8} = -\frac{1}{2} \]

Conclusion

The equation \(4x^2 - 4x - 3 = 0\) has two real solutions: \(x = \frac{3}{2}\) and \(x = -\frac{1}{2}\). Therefore, Joana's assertion that the quadratic has no solutions is incorrect.

The correct response is: No, because she did not factor out the GCF from the first grouping. This led to not realizing that the equation still has solutions.