Determine whether the values −1 and 7/3 are solutions to the quadratic equation 3xexponent2−4x−4=3 .(1 point)

Responses

Only x= 7/3 is a solution.
Only x equals Start Fraction 7 over 3 End Fraction is a solution.

Only x= −1 is a solution.
Only x equals negative 1 is a solution.

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

Both x=−1 and x= 7/3 are solutions.

1 answer

To determine if the values \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the equation \( 3x^2 - 4x - 4 = 3 \), we need to substitute these values into the equation and see if they satisfy it.

First, simplify the equation:

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

Now, we will substitute \( x = -1 \):

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

Since this simplifies to \( 0 = 0 \), \( x = -1 \) is a solution.

Next, we will substitute \( x = \frac{7}{3} \):

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

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

\[ \left(\frac{7}{3}\right)^2 = \frac{49}{9} \]

Now substitute this into the equation:

\[ 3 \cdot \frac{49}{9} - 4 \cdot \frac{7}{3} - 7 = \frac{147}{9} - \frac{28}{3} - 7 \]

Convert \(-\frac{28}{3}\) and \(-7\) to fractions with a denominator of 9:

\[ -\frac{28}{3} = -\frac{84}{9} \quad \text{and} \quad -7 = -\frac{63}{9} \]

Now, combine all terms:

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

Since this simplifies to \( 0 = 0 \), \( x = \frac{7}{3} \) is also a solution.

Therefore, both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the equation.

The answer is: Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions.