### Discussion Board Rubric

#### Prompt:
Amy and Richard each solved an equation using the quadratic formula.

#### Amy's Equation and Method
[tex]\[ 4x^2 + 7x - 20 = 0 \][/tex]

Step 1: [tex]\[ x = \frac{-7 \pm \sqrt{7^2 - 4(4)(-20)}}{2(4)} \][/tex]

Step 2: [tex]\[ x = \frac{-7 \pm \sqrt{49 + 320}}{8} \][/tex]

Step 3: [tex]\[ x = \frac{-7 \pm \sqrt{369}}{8} \][/tex]

#### Richard's Equation and Method
[tex]\[ x^2 + 6x + 8 = 0 \][/tex]

Step 1: [tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(8)}}{2(1)} \][/tex]

Step 2: [tex]\[ x = \frac{-6 \pm \sqrt{36 - 32}}{2} \][/tex]

Step 3: [tex]\[ x = \frac{-6 \pm \sqrt{4}}{2} \][/tex]

Step 4: [tex]\[ x = \frac{-6 \pm 2}{2} \][/tex]

Step 5: [tex]\[ x = -3 \pm 1 \][/tex]

Step 6: [tex]\[ x = -4 \text{ or } x = -2 \][/tex]

Note: The prompt has been corrected to properly use the quadratic formula steps and corrected any errors in the original problem setup and solution steps.



Answer :

Sure! Let's carefully analyze how Amy and Richard solved their respective quadratic equations using the quadratic formula and address any errors they made along the way.

### Amy's Approach:
Let's start with Amy's equation:

[tex]\[ 4x^2 + 7x - 20 = 0 \][/tex]

Step 1: Apply the Quadratic Formula

The quadratic formula is given by:

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

For Amy's equation, [tex]\(a = 4\)[/tex], [tex]\(b = 7\)[/tex], and [tex]\(c = -20\)[/tex]. Applying these values, we get:

[tex]\[ x = \frac{-7 \pm \sqrt{7^2 - 4(4)(-20)}}{2(4)} \][/tex]

Step 2: Simplify the terms inside the square root

First, calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):

[tex]\[b^2 = 7^2 = 49\][/tex]
[tex]\[4ac = 4 \cdot 4 \cdot -20 = -320\][/tex]
[tex]\[b^2 - 4ac = 49 + 320 = 369\][/tex]

So the expression inside the square root is [tex]\( \sqrt{369} \)[/tex]. Now our equation looks like this:

[tex]\[ x = \frac{-7 \pm \sqrt{369}}{8} \][/tex]

Step 3: Identify the roots in simplified form

Thus, the roots of Amy's quadratic equation are:

[tex]\[ x = \frac{-7 + \sqrt{369}}{8} \quad \text{and} \quad x = \frac{-7 - \sqrt{369}}{8} \][/tex]

So, Amy's mistake was in her step 2; she mistakenly calculated [tex]\(49 - 80\)[/tex] instead of getting [tex]\(49 + 320\)[/tex]. The correct form of her solution is:

[tex]\[ x = \frac{-7 \pm \sqrt{369}}{8} \][/tex]

### Richard's Approach:
Let's now analyze the equation and steps provided by Richard. He solved the quadratic equation:

[tex]\[ x^2 + 6x + 8 = 0 \][/tex]

Step 1: Apply the Quadratic Formula

Using the quadratic formula, with [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = 8\)[/tex]:

[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(8)}}{2(1)} \][/tex]

Step 2: Simplify the terms inside the square root

First, calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):

[tex]\[b^2 = 6^2 = 36\][/tex]
[tex]\[4ac = 4 \cdot 1 \cdot 8 = 32\][/tex]
[tex]\[b^2 - 4ac = 36 - 32 = 4\][/tex]

So, the expression inside the square root is [tex]\( \sqrt{4} \)[/tex]. Now our equation looks like this:

[tex]\[ x = \frac{-6 \pm \sqrt{4}}{2} \][/tex]

Step 3: Simplify the square root and find the roots

[tex]\[ \sqrt{4} = 2 \][/tex]

So:

[tex]\[ x = \frac{-6 \pm 2}{2} \][/tex]

Breaking this down, we get two solutions:

[tex]\[ x_1 = \frac{-6 + 2}{2} = \frac{-4}{2} = -2 \][/tex]
[tex]\[ x_2 = \frac{-6 - 2}{2} = \frac{-8}{2} = -4 \][/tex]

So Richard's solution gives the roots [tex]\(x = -2\)[/tex] and [tex]\(x = -4\)[/tex]. However, there seems to be a discrepancy previous stating that his steps would result in `root1: 4.0, root2: 2.0`, indicating Richard's initial equation may require clarification on variables or initial condition's (misinterpreted integer signs possibly).

Therefore, Richard's final roots from accurate steps are:
[tex]\[ x_1 = -2 \][/tex]
[tex]\[ x_2 = -4 \][/tex]

In summary:
- Amy corrected calculation should yield roots about [tex]\(x = \frac{-7 \pm \sqrt{369}}{8}\)[/tex]
- Richard’s quadratic after accurate discrimination step finally calculated [tex]\(x_1 = 4 \)[/tex], [tex]\(x_2 = 2\)[/tex] should re-evaluated relation consistent context.

However, each took necessary steps employing Quadratic definition efficiently!