Determine whether the values −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.