10.

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 to 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 -12 that add up to -4.

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

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 did not factor out the GCF from the first grouping.

1 answer

Let's analyze Joana's conclusion based on her steps for solving the quadratic equation \(4x^2 - 4x - 3 = 0\).

Step 1: Joana rewrote the middle term, splitting \(-4x\) into \(2x\) and \(-6x\), which is correct since \(2x - 6x = -4x\). The equation now reads:

\[ 4x^2 + 2x - 6x - 3 = 0 \]

Step 2: She grouped the terms:

\[ (4x^2 + 2x) + (-6x - 3) = 0 \]

Step 3: She factored each grouping:

\[ x(4x + 2) - 3(2x + 1) = 0 \]

At this point, Joana noticed that the two groupings \(x(4x + 2)\) and \(-3(2x + 1)\) do not share a common binomial factor.

Joana's conclusion that the quadratic equation has no solutions is not correct. The lack of a common factor in the groupings does not imply that there are no solutions to the quadratic equation.

Quadratic equations can still have solutions even if they don't factor neatly. The equation can also be solved using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For \(4x^2 - 4x - 3 = 0\), we can take \(a = 4\), \(b = -4\), and \(c = -3\):

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-3)}}{2(4)} \] \[ x = \frac{4 \pm \sqrt{16 + 48}}{8} \] \[ x = \frac{4 \pm \sqrt{64}}{8} \] \[ x = \frac{4 \pm 8}{8} \]

From this, we find two possible solutions:

  1. \(x = \frac{12}{8} = \frac{3}{2}\)
  2. \(x = \frac{-4}{8} = -\frac{1}{2}\)

So, the correct response about Joana's conclusion is:

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