Question

Determine whether the values -1 and 7/3 are solutions to the quadratic equation 3x^2-4x-4=3
A. Neither x= -1 nor x= 7/3 is a solution
B. Only x = -1 is a solution
C. Only x = 7/3 is a solution
D. Both x = -1 and x=7/3 are solutions

Answer 1

To determine whether -1 and \( \frac{7}{3} \) are solutions to the equation \( 3x^2 - 4x - 4 = 3 \), we first simplify the equation.

Subtract 3 from both sides:

\[
3x^2 - 4x - 4 - 3 = 0
\]
\[
3x^2 - 4x - 7 = 0
\]

Now, let's check each value:

1. **Checking \( x = -1 \)**:

\[
3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0
\]

The left side equals 0, so \( x = -1 \) is a solution.

2. **Checking \( x = \frac{7}{3} \)**:

Substituting \( x = \frac{7}{3} \):

\[
3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7
\]

Calculating \( 3\left(\frac{7}{3}\right)^2 \):

\[
3 \cdot \frac{49}{9} = \frac{147}{9}
\]

Calculating \( -4\left(\frac{7}{3}\right) \):

\[
-\frac{28}{3} = -\frac{84}{9}
\]

Now we put it all together:

\[
\frac{147}{9} - \frac{84}{9} - 7 = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0
\]

The left side also equals 0, so \( x = \frac{7}{3} \) is a solution.

Since both \( x = -1 \) and \( x = \frac{7}{3} \) satisfy the equation, the answer is:

**D. Both x = -1 and x = 7/3 are solutions.**