Question

Determine whether the values −1
−
1
and 73
7
3
are solutions to the quadratic equation 3x2−4x−4=3
3
𝑥
2
−
4
𝑥
−
4
=
3
.(1 point)
Responses

Neither x=−1
𝑥
=
−
1
nor x=73
𝑥
=
7
3
is a solution.
Neither x equals negative 1 nor x equals Start Fraction 7 over 3 End Fraction is a solution.

Only x=−1
𝑥
=
−

Answer 1

To determine whether the values \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \), we'll need to substitute each value into the left side of the equation and check if it equals 3.

### First, simplify the equation:
We can rewrite the equation as follows:
\[
3x^2 - 4x - 4 - 3 = 0
\]
which simplifies to:
\[
3x^2 - 4x - 7 = 0
\]

### Testing \( x = -1 \):
Substituting \( x = -1 \):
\[
3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0
\]
So, \( 3(-1)^2 - 4(-1) - 7 = 0 \), which means \( x = -1 \) is **not** a solution to the equation \( 3x^2 - 4x - 7 = 0 \).

### Testing \( x = \frac{7}{3} \):
Substituting \( x = \frac{7}{3} \):
\[
3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7
\]
Calculating \( \left(\frac{7}{3}\right)^2 \):
\[
\left(\frac{7}{3}\right)^2 = \frac{49}{9}
\]
Substituting back in:
\[
3 \times \frac{49}{9} - 4 \times \frac{7}{3} - 7 = \frac{147}{9} - \frac{28}{3} - 7
\]
Converting \( \frac{28}{3} \) and \( 7 \) to fractions over 9:
\[
\frac{28}{3} = \frac{84}{9} \quad \text{and} \quad 7 = \frac{63}{9}
\]
Now substituting those values in:
\[
\frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0
\]
So, \( 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 = 0 \), which means \( x = \frac{7}{3} \) is also **not** a solution to the equation.

### Conclusion:
Both \( x = -1 \) and \( x = \frac{7}{3} \) are **not** solutions to the equation \( 3x^2 - 4x - 7 = 0 \).

Thus, the answer is:
**Neither \( x = -1 \) nor \( x = \frac{7}{3} \) is a solution.**