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=0 Step 2: (4x2+2x)+(−6x−3)=0 Step 3: x(4x+2)+(−3(2x+1))=0 (1 point) Responses 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 . 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.

Answer 1

Joana's conclusion that the quadratic equation \( 4x^2 - 4x - 3 = 0 \) has no solutions is incorrect. Let's break down the reasoning.

1. Review of the Quadratic Equation: The equation \( 4x^2 - 4x - 3 = 0 \) can be solved using various methods, including factoring, the quadratic formula, or completing the square. The presence of a common binomial in grouping is not a necessary condition to derive solutions for a quadratic equation.

2. Step Analysis:

   • Step 1: Joana rewrites the middle term as \( 2x - 6x \): \( 4x^2 + 2x - 6x - 3 = 0 \).
   • Step 2: She groups it as \( (4x^2 + 2x) + (-6x - 3) = 0 \).
   • Step 3: Joana tried factoring each group: \( x(4x + 2) - 3(2x + 1) = 0 \).

   At this stage, it's clear that the common factors or solutions may not be obvious, but that doesn't mean the equation lacks solutions.

3. Factoring or Finding Roots: Joana should have moved forward with either using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) or looking for factors that sum and multiply appropriately, instead of stopping at the inability to find a shared binomial.

Given these considerations, the correct response to Joana’s conclusion is:

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

The equation \( 4x^2 - 4x - 3 = 0 \) does indeed have solutions, and continuing with the quadratic formula would provide those solutions.