To determine if 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 by subtracting 3 from both sides:
\[ 3x^2 - 4x - 4 - 3 = 0 \] which simplifies to: \[ 3x^2 - 4x - 7 = 0 \]
Now we will check each value:
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For \( x = -1 \): \[ 3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0 \] Since the left side equals the right side (0 = 0), \( x = -1 \) is a solution.
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For \( x = \frac{7}{3} \): \[ 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 \] Calculating \( 3\left(\frac{7}{3}\right)^2 \): \[ 3\left(\frac{49}{9}\right) = \frac{147}{9} \]
Now calculate \( -4\left(\frac{7}{3}\right) \): \[ -4 \cdot \frac{7}{3} = -\frac{28}{3} \]
To combine the terms, we convert \( -\frac{28}{3} \) to have a common denominator: \[ -\frac{28}{3} = -\frac{84}{9} \]
Now substitute back into the equation: \[ \frac{147}{9} - \frac{84}{9} - 7 = \frac{147 - 84}{9} - 7 = \frac{63}{9} - 7 \] \[ = \frac{63}{9} - \frac{63}{9} = 0 \] Since the left side equals the right side (0 = 0), \( x = \frac{7}{3} \) is also a solution.
Therefore, the final answer is True: both -1 and \( \frac{7}{3} \) are solutions to the given quadratic equation.
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