To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 15 \), and \( c = 36 \).
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Calculate the discriminant: \[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]
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Substitute values into the quadratic formula: \[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2} \]
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Calculate the two possible solutions: \[ x_1 = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \] \[ x_2 = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
The zeros of the function are \( x = -12 \) and \( x = -3 \).
When listing the smaller zero and greater zero:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
So the final answer is:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)