Question

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.

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.