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.
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Answer 1

To analyze Joana's conclusion about the quadratic equation \(4x^2 - 4x - 3 = 0\), let's review her steps:

1. Step 1: Joana rewrites the equation as \(4x^2 + 2x - 6x - 3 = 0\). This is a valid step, as the \( -4x \) can be split into \(2x\) and \(-6x\).

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

3. Step 3: Joana attempts to factor by grouping: \(x(4x + 2) - 3(2x + 1) = 0\).

In her third step, Joana finds that the two groupings, \(x(4x + 2)\) and \(-3(2x + 1)\), do not have a common binomial factor.

Now, examining her conclusion:

• It is incorrect to state that the equation has no solutions simply because her grouping technique did not yield a common binomial. The quadratic equation can still be solved using other methods, such as using the quadratic formula or completing the square.

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

Joana's conclusion that the quadratic equation has no solutions is not valid based on her method alone. Thus, she can explore more options to solve the equation.