Question

Joana solved the equation 4x2−4x−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 has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question. Step 1: 4x2+2x+(−6x)−3=0Step 2: (4x2+2x)+(−6x−3)=0Step 3: x(4x+2)+(−3(2x+1))=0(1 point)ResponsesNo, 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 negative 12 that add up to negative 4 .Yes, because the two groupings do not share a common binomial.

Answer 1

No — her conclusion is incorrect.

Error: in the first grouping 4x^2 + 2x she factored out x to get x(4x+2), but the greatest common factor is 2x. Factoring correctly:

4x^2 + 2x = 2x(2x+1)
−6x − 3 = −3(2x+1)

Now the common binomial (2x+1) can be factored:

4x^2 − 4x − 3 = 2x(2x+1) − 3(2x+1) = (2x+1)(2x−3).

So the equation (2x+1)(2x−3)=0 gives x = −1/2 or x = 3/2.