solution sets of quadratic equations quick check
determine whether the values -1 and $\frac{7}{3}$ are solutions to the quadratic equation $3x^2 - 4x - 4 = 3$. (1 point)
$\bigcirc$ only $x = \frac{7}{3}$ is a solution.
$\bigcirc$ only $x = -1$ is a solution.
$\bigcirc$ both $x = -1$ and $x = \frac{7}{3}$ are solutions.
$\bigcirc$ neither $x = -1$ nor $x = \frac{7}{3}$ is a solution.

Question

Accepted Answer

# Explanation:
## Step1: Check $ x = -1 $
Substitute $ x = -1 $ into $ 3x^2 - 4x - 4 $:
$ 3(-1)^2 - 4(-1) - 4 = 3(1) + 4 - 4 = 3 + 0 = 3 $
Which equals the right - hand side (3), so $ x=-1 $ is a solution.

## Step2: Check $ x=\frac{7}{3} $
Substitute $ x = \frac{7}{3} $ into $ 3x^2 - 4x - 4 $:
$$
\begin{align*}
3\times(\frac{7}{3})^2-4\times\frac{7}{3}-4&=3\times\frac{49}{9}-\frac{28}{3}-4\
&=\frac{49}{3}-\frac{28}{3}-4\
&=\frac{49 - 28}{3}-4\
&=\frac{21}{3}-4\
&=7 - 4\
&=3
\end{align*}
$$
Which equals the right - hand side (3), so $ x = \frac{7}{3} $ is a solution.

# Answer:
Both $ x=-1 $ and $ x = \frac{7}{3} $ are solutions.

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QUESTION IMAGE

Question

Explanation:

Step1: Check \( x = -1 \)

Step2: Check \( x=\frac{7}{3} \)

Answer: