Determine whether the values −1 and 7/3 are solutions to the quadratic equation 3x^2−4x−4=3.

1 answer

To determine whether the values \( -1 \) and \( \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \), we first rewrite the equation in standard form. We subtract 3 from both sides of the equation:

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

This simplifies to:

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

Now, we will test each value.

Testing \( x = -1 \):

Substituting \( x = -1 \) into the equation:

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

Calculating the left side:

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

Since the left side equals 0, \( x = -1 \) is indeed a solution.

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

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

\[ 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 = 0 \]

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

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

Now substitute it into the equation:

\[ 3 \cdot \frac{49}{9} - 4 \cdot \frac{7}{3} - 7 = 0 \]

Calculating \( 3 \cdot \frac{49}{9} \):

\[ \frac{147}{9} \]

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

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

Now we can combine all parts:

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

Combining the fractions:

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

Thus, \( x = \frac{7}{3} \) is also a solution.

In conclusion, both \( -1 \) and \( \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \).