To determine the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \), we will set the equation equal to zero and solve for \( x \):
\[ 2x^2 + 9x + 4 = 0 \]
To solve this quadratic equation, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \( a = 2 \), \( b = 9 \), and \( c = 4 \). Now we will calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 9^2 - 4 \times 2 \times 4 \] \[ = 81 - 32 \] \[ = 49 \]
Now, we can substitute the values of \( a \), \( b \), and the discriminant into the quadratic formula:
\[ x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2} \] \[ = \frac{-9 \pm 7}{4} \]
Now we will calculate the two possible values for \( x \):
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For \( -9 + 7 \): \[ x = \frac{-2}{4} = -\frac{1}{2} \]
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For \( -9 - 7 \): \[ x = \frac{-16}{4} = -4 \]
Thus, the zeros of the quadratic function \( y = 2x^2 + 9x + 4 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -4 \]