Question

Joana solved the equation 4x2−4x−3=0
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

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

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 groupinds have no shared binomial.

Answer 1

Joana's conclusion that the quadratic equation \(4x^2 - 4x - 3 = 0\) has no solutions is incorrect. We need to analyze her reasoning and the work she completed.

In Step 1, Joana rewrote the quadratic equation by splitting the middle term, \( -4x \), into \( +2x \) and \( -6x \). This is a proper technique called "splitting the middle term" for factoring, although it might not lead to the simplest or most straightforward grouping.

In Step 2, she grouped the terms as \( (4x^2 + 2x) + (-6x - 3) = 0 \) and then factored out terms from each grouping in Step 3, which resulted in \( x(4x + 2) - 3(2x + 1) = 0 \). The two groupings did not share a common binomial, which made it impossible for her to factor it further to find a common factor.

However, the fact that the groupings did not share a common factor does not mean that the quadratic has no solutions. She can use other methods to solve the quadratic equation, such as the quadratic formula:

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

For the equation \( 4x^2 - 4x - 3 = 0 \):
- \( a = 4 \)
- \( b = -4 \)
- \( c = -3 \)

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

Since the discriminant is positive (\(64 > 0\)), there are two real solutions.

So, the correct response is:

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