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

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 evaluate Joana's conclusion regarding the quadratic equation \(4x^2 - 4x - 3 = 0\), we need to review her steps and understand the reasoning behind them.

1. Rewriting the equation: In her Step 1, Joana correctly rewrites the equation as \(4x^2 + 2x - 6x - 3 = 0\). She is breaking down the middle term \(-4x\) into \(2x\) and \(-6x\).

2. Grouping terms: In Step 2, she groups the terms into two parts: \((4x^2 + 2x) + (-6x - 3) = 0\). This is a correct grouping.

3. Factoring: In Step 3, she factors out the GCF from the first grouping \(x(4x + 2)\) and incorrectly attempts to factor out from the second group. Her approach includes \(-3(2x + 1)\), which is not factored correctly.

It seems Joana misinterpreted the grouping process. For proper factorization, she should have factored each grouping correctly. Specifically, the first group could become \(2x(2x + 1)\), and the second group could be made consistent with the terms of the first, but this was not the approach she took.

Despite complications in factoring, the conclusion that the quadratic has no solutions is not correct. Quadratic equations can have solutions even if they cannot be factored neatly into shared binomials. The discriminant can help us determine the nature of the roots:

Calculating the discriminant \(b^2 - 4ac\):

• Here \(a = 4\), \(b = -4\), and \(c = -3\).
• The discriminant is \((-4)^2 - 4(4)(-3) = 16 + 48 = 64\).

Since the discriminant is positive, this indicates that there are two real solutions.

Thus, the correct response is:

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