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Question

Question 1
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

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.

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 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.
Yes, because the two groupings do not share a common binomial.
Question 2
11.

Each of the following quadratic equations can be solved by factoring. Which one has only one distinct solution?

(1 point)
Responses

x2 − 5x + 4 = 0
x2 − 5x + 4 = 0

x2 + 4x + 4 = 0
x2 + 4x + 4 = 0

x2 + 5x + 4 = 0
x2 + 5x + 4 = 0

x2 − 4 = 0

Answer 1

Question 1

Joana's conclusion is not correct. She states:

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

In this case, Joana can still find the roots of the equation despite the lack of a common binomial after her grouping steps.

Question 2

To determine which quadratic equation has only one distinct solution, we need to find the equations that can be factored into a perfect square trinomial or an irreducible factor.

1. \( x^2 - 5x + 4 = 0 \)

   • Factors as \( (x - 1)(x - 4) = 0 \) → Solutions: \( x = 1, x = 4 \) (two distinct solutions).

2. \( x^2 + 4x + 4 = 0 \)

   • Factors as \( (x + 2)(x + 2) = 0 \) → Solution: \( x = -2 \) (one distinct solution).

3. \( x^2 + 5x + 4 = 0 \)

   • Factors as \( (x + 1)(x + 4) = 0 \) → Solutions: \( x = -1, x = -4 \) (two distinct solutions).

4. \( x^2 - 4 = 0 \)

   • Factors as \( (x - 2)(x + 2) = 0 \) → Solutions: \( x = 2, x = -2 \) (two distinct solutions).

Thus, the equation that has only one distinct solution is:

• \( x^2 + 4x + 4 = 0 \).