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 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.
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
−
12
that add up to −4
−
4
.

Answer 1

Joana’s conclusion is not correct because she prematurely concluded that the quadratic equation has no solutions based on her attempt to group and factor.

The relevant option is:

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

In her work, even though the two groupings do not share a common binomial, it is still possible to solve the quadratic equation through other methods, such as using the quadratic formula or completing the square. The equation \(4x^2 - 4x - 3 = 0\) may have solutions even if Joana's method of grouping did not yield any factors.

To confirm this, we can apply 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:

\[ b^2 - 4ac = (-4)^2 - 4(4)(-3) = 16 + 48 = 64 \]

Since the discriminant is positive, there are two real solutions. Thus, Joana's conclusion that the equation has no solutions is incorrect.