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.
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 she did not find a common binomial factor in her grouping does not mean that the quadratic equation has no solutions.

Here's why:

1. Reviewing her steps:

   • In Step 1, she correctly rewrote the middle term $-4x$ as $2x-6x$.
   • In Step 2, she grouped the terms as $(4x^2+2x)+(-6x-3)=0$.
   • However, in Step 3, her factoring is not accurate. The correct grouping and factoring should lead to finding common factors.

2. Finding solutions:

   • If the grouping fails to yield a common binomial factor, it doesn't imply no solutions exist; it merely indicates that this method of factoring might not work. The quadratic formula could still provide solutions, regardless of whether she was able to factor it through grouping.

3. Using the Quadratic Formula:

   • The equation $4x^2-4x-3=0$ can be solved using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. For this equation:
     • $a=4$
     • $b=-4$
     • $c=-3$
     • The discriminant $b^2-4ac=(-4{)}^2-4(4)(-3)=16+48=64$, which is positive, indicating that there are two real solutions.

Therefore, the correct response is: No, because she can still continue the steps even if the two groupings have no shared binomial.