Determine the zeros of the quadratic function \(y=x^2+15x+36\). (When listing the smaller

Question

zero, remember that negative values are always smaller than positive values.) (1 point)

smaller zero: \(x=\) ; greater zero: \(x=\)

Answer 1

To find the zeros of the quadratic function \(y = x^2 + 15x + 36\), we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, the coefficients are:

• \(a = 1\)
• \(b = 15\)
• \(c = 36\)

We first need to compute the discriminant \((b^2 - 4ac)\):

\[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]

Now we can substitute the values into the quadratic formula:

\[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} \]

Since \(\sqrt{81} = 9\), we have:

\[ x = \frac{-15 \pm 9}{2} \]

This gives us two possible solutions:

1. For \(x = \frac{-15 + 9}{2}\):

\[ x = \frac{-6}{2} = -3 \]

2. For \(x = \frac{-15 - 9}{2}\):

\[ x = \frac{-24}{2} = -12 \]

Thus, the zeros of the quadratic function are:

• Smaller zero: \(x = -12\)
• Greater zero: \(x = -3\)

To summarize:

smaller zero: \(x = -12\) ; greater zero: \(x = -3\)