Question

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

Answer 1

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 \( 3x^2 - 4x - 4 = 3 \).