To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
where \( a = 1 \), \( b = 15 \), 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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Since the discriminant is positive, we have two distinct real roots. Now we calculate the zeros: \[ x = \frac{{-15 \pm \sqrt{81}}}{2 \cdot 1} \] \[ x = \frac{{-15 \pm 9}}{2} \]
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Calculate the two solutions:
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First solution (smaller zero): \[ x = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12 \]
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Second solution (greater zero): \[ x = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3 \]
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Thus, the zeros of the function are:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
In summary:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)
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