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Joana solve the equation 4x^2 - 4x - 3 = 0. she followed the steps to solve a quadratic equation by grouping. However, institution is that the two groupings are not share common binomial, as shown in her work below. she concluded that the quadratic equation has no solution solutions. is Joana’s conclusion correct? Review Joana’s work to answer the question m.

Step 1: 4x^2 + 2x + (-6x) - 3 = 0
Step 2: (4x^2 + 2x) + (-6x - 3) = 0
Step 3: x(4x + 2x) + (-3(2x + 1))= 0
a no, because she did not factor out the GCF from the first grouping
b no, because she can still continue the steps, even if the two groupings have no shared binomial
c yes, because there are no factors of -12 that add up to -4
d yes, because the two groupings do not share a common binomial

1 answer

To determine whether Joana's conclusion is correct, let's analyze the steps she took in solving the quadratic equation \(4x^2 - 4x - 3 = 0\).

  1. Step 1: She rewrote the equation as \(4x^2 + 2x - 6x - 3 = 0\). This step is valid as \(2x - 6x = -4x\).

  2. Step 2: In this step, she grouped the terms as \((4x^2 + 2x) + (-6x - 3) = 0\).

  3. Step 3: She attempted to factor each grouping separately. Her factorization is:

    • From the first grouping: \(4x^2 + 2x\) could be factored as \(2x(2x + 1)\).
    • From the second grouping: \(-6x - 3\) can be factored as \(-3(2x + 1)\).

At this point, Joana should have factored the expression as follows: \[ 2x(2x + 1) - 3(2x + 1) = 0 \] This shows a common binomial factor of \(2x + 1\). Consequently, she could factor the entire expression as: \[ (2x + 1)(2x - 3) = 0 \]

Since the quadratic equation can be factored, the conclusion that the equation has no solutions is incorrect.

Given the options, let's evaluate them:

a) No, because she did not factor out the GCF from the first grouping.
This statement is somewhat misleading. While she didn't factor correctly, the overall reasoning about the lack of solutions is the main issue.

b) No, because she can still continue the steps, even if the two groupings have no shared binomial.
This is correct. She could still have continued and identified a solution.

c) Yes, because there are no factors of -12 that add up to -4.
This is incorrect as it doesn't apply to her method or the current context.

d) Yes, because the two groupings do not share a common binomial.
While it is true that she might not have seen a common binomial, this does not negate the possibility of finding a solution.

From this analysis, the best answer is:

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

Joana's conclusion is incorrect because she can further simplify and solve the equation properly.