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!
### 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!
