To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)
Now, we will calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 \] \[ = 225 - 144 \] \[ = 81 \]
Since the discriminant is positive, there are two distinct real zeros.
Next, we will apply the quadratic formula:
\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} \] \[ = \frac{-15 \pm 9}{2} \]
Now we will calculate the two possible values for \( x \):
-
For the plus sign: \[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
-
For the minus sign: \[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]
Now we can list the zeros. The smaller zero is \( -12 \) and the greater zero is \( -3 \).
So, we have:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)
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