To determine if the values \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \), we need to substitute each value into the equation and see if the equation holds true.
First, let's simplify the equation:
\[ 3x^2 - 4x - 4 - 3 = 0 \]
This simplifies to:
\[ 3x^2 - 4x - 7 = 0 \]
Now, let's check each potential solution.
1. For \( x = -1 \):
\[ 3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0 \]
So, \( x = -1 \) is a solution.
2. For \( x = \frac{7}{3} \):
\[ 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 \]
Calculating \( 3\left(\frac{7}{3}\right)^2 \):
\[ 3\left(\frac{49}{9}\right) = \frac{147}{9} \]
Calculating \( 4\left(\frac{7}{3}\right) = \frac{28}{3} = \frac{84}{9} \).
Now, putting it all together:
\[ \frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0 \]
So, \( x = \frac{7}{3} \) is also a solution.
Since both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the equation, the correct response is:
Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions.
1 answer