Determine whether the values −1

Question

−
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

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

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.

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

Answer 1

To determine whether \( x = -1 \) and \( x = \frac{7}{3} \) are solutions to the quadratic equation \( 3x^2 - 4x - 4 = 3 \), we can substitute each value into the equation and see if the left side equals the right side (which is 3).

1. Testing \( x = -1 \):

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

   Since the left-hand side equals the right-hand side (3 = 3), \( x = -1 \) is a solution.

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

   \[ 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 4 \] First, calculate \(\left(\frac{7}{3}\right)^2 = \frac{49}{9}\):

   \[ 3\left(\frac{49}{9}\right) - 4\left(\frac{7}{3}\right) - 4 = \frac{147}{9} - \frac{28}{3} - 4 \] Convert \( -\frac{28}{3} \) and \( -4 \) to fractions with a denominator of 9:

   \[ -\frac{28}{3} = -\frac{84}{9}, \quad -4 = -\frac{36}{9} \]

   Now substituting these back:

   \[ \frac{147}{9} - \frac{84}{9} - \frac{36}{9} = \frac{147 - 84 - 36}{9} = \frac{27}{9} = 3 \]

   Since the left-hand side also equals the right-hand side (3 = 3), \( x = \frac{7}{3} \) is also a solution.

Conclusion: Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions.

So the correct response is:

Both \( x = -1 \) and \( x = \frac{7}{3} \) are solutions.