Question

Determine whether the values −1
−
1
and 73
7
3
are solutions to the quadratic equation 3x2−4x−4=3
3
𝑥
2
−
4
𝑥
−
4
=
3
.(1 point)
Responses

Neither x=−1
𝑥
=
−
1
nor x=73
𝑥
=
7
3
is a solution.
Neither x equals negative 1 nor x equals Start Fraction 7 over 3 End Fraction is a solution.

Only x=73
𝑥
=
7
3
is a solution.
Only x equals Start Fraction 7 over 3 End Fraction is a solution.

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

Both x=−1
𝑥
=
−
1
and x=73
𝑥
=
7
3
are solutions.

Answer 1

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.**