. 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 has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.
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
(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.
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 that add up to −4.
Yes, because there are no factors of −12 that add up to −4.
Yes, because the two groupings do not share a common binomial.
1 answer
No, because she can still continue the steps even if the two groupings have no shared binomial.
Joana's conclusion is incorrect. Even if the two groupings do not share a common binomial, she can still find solutions to the quadratic equation by considering other factoring methods or the quadratic formula. The presence of shared binomials can help simplify the process, but their absence does not mean that the quadratic equation has no solutions.
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