Answer :
To determine the original equation Michele solved, we need to determine the roots derived from the quadratic formula:
[tex]\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 7 \cdot (-2)}}{14} \][/tex]
Start by simplifying inside the square root:
[tex]\[ x = \frac{5 \pm \sqrt{25 + 56}}{14} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{81}}{14} \][/tex]
Since [tex]\(\sqrt{81} = 9\)[/tex]:
[tex]\[ x = \frac{5 \pm 9}{14} \][/tex]
This results in two solutions:
[tex]\[ x_1 = \frac{5 + 9}{14} = \frac{14}{14} = 1 \][/tex]
[tex]\[ x_2 = \frac{5 - 9}{14} = \frac{-4}{14} = -\frac{2}{7} \][/tex]
Now, we need to check which equation these roots satisfy among the given choices:
### Option A: [tex]\(7x^2 - 5x - 2 = -1\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) - 2 = 7 - 5 - 2 = 0 \neq -1 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) - 2 = 7\left(\frac{4}{49}\right) + \frac{10}{7} - 2 = \frac{28}{49} + \frac{70}{49} - \frac{98}{49} = 0 \neq -1 \][/tex]
Option A is not correct.
### Option B: [tex]\(7x^2 - 5x + 6 = -8\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) + 6 = 7 - 5 + 6 = 8 \neq -8 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) + 6 = 7\left(\frac{4}{49}\right) + \frac{10}{7} + 6 = \frac{28}{49} + \frac{70}{49} + \frac{294}{49} = \frac{392}{49} = 8 \neq -8 \][/tex]
Option B is not correct.
### Option C: [tex]\(7x^2 - 5x + 3 = 5\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) + 3 = 7 - 5 + 3 = 5 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) + 3 = 7\left(\frac{4}{49}\right) + \frac{10}{7} + 3 = \frac{28}{49} + \frac{70}{49} + \frac{147}{49} = \frac{245}{49} = 5 \][/tex]
Both roots satisfy the equation.
### Option D: [tex]\(7x^2 - 5x + 5 = 3\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) + 5 = 7 - 5 + 5 = 7 \neq 3 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) + 5 = 7\left(\frac{4}{49}\right) + \frac{10}{7} + 5 = \frac{28}{49} + \frac{70}{49} + \frac{245}{49} = \frac{343}{49} = 7 \neq 3 \][/tex]
Option D is not correct.
Therefore, the correct answer is [tex]\(\boxed{C}\)[/tex].
[tex]\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 7 \cdot (-2)}}{14} \][/tex]
Start by simplifying inside the square root:
[tex]\[ x = \frac{5 \pm \sqrt{25 + 56}}{14} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{81}}{14} \][/tex]
Since [tex]\(\sqrt{81} = 9\)[/tex]:
[tex]\[ x = \frac{5 \pm 9}{14} \][/tex]
This results in two solutions:
[tex]\[ x_1 = \frac{5 + 9}{14} = \frac{14}{14} = 1 \][/tex]
[tex]\[ x_2 = \frac{5 - 9}{14} = \frac{-4}{14} = -\frac{2}{7} \][/tex]
Now, we need to check which equation these roots satisfy among the given choices:
### Option A: [tex]\(7x^2 - 5x - 2 = -1\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) - 2 = 7 - 5 - 2 = 0 \neq -1 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) - 2 = 7\left(\frac{4}{49}\right) + \frac{10}{7} - 2 = \frac{28}{49} + \frac{70}{49} - \frac{98}{49} = 0 \neq -1 \][/tex]
Option A is not correct.
### Option B: [tex]\(7x^2 - 5x + 6 = -8\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) + 6 = 7 - 5 + 6 = 8 \neq -8 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) + 6 = 7\left(\frac{4}{49}\right) + \frac{10}{7} + 6 = \frac{28}{49} + \frac{70}{49} + \frac{294}{49} = \frac{392}{49} = 8 \neq -8 \][/tex]
Option B is not correct.
### Option C: [tex]\(7x^2 - 5x + 3 = 5\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) + 3 = 7 - 5 + 3 = 5 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) + 3 = 7\left(\frac{4}{49}\right) + \frac{10}{7} + 3 = \frac{28}{49} + \frac{70}{49} + \frac{147}{49} = \frac{245}{49} = 5 \][/tex]
Both roots satisfy the equation.
### Option D: [tex]\(7x^2 - 5x + 5 = 3\)[/tex]
Replace [tex]\(x\)[/tex] with 1:
[tex]\[ 7(1)^2 - 5(1) + 5 = 7 - 5 + 5 = 7 \neq 3 \][/tex]
Replace [tex]\(x\)[/tex] with [tex]\(-\frac{2}{7}\)[/tex]:
[tex]\[ 7\left(-\frac{2}{7}\right)^2 - 5\left(-\frac{2}{7}\right) + 5 = 7\left(\frac{4}{49}\right) + \frac{10}{7} + 5 = \frac{28}{49} + \frac{70}{49} + \frac{245}{49} = \frac{343}{49} = 7 \neq 3 \][/tex]
Option D is not correct.
Therefore, the correct answer is [tex]\(\boxed{C}\)[/tex].
