Question

Find the zeros of the quadratic equation: x^2−16x−13=4
x
2
−
16
x
−
13
=
4
Enter the smaller of the 2 values first.

(1 point)
x=
x
=

, x=
x
=

Answer 1

To find the zeros of the quadratic equation \( x^2 - 16x - 13 = 4 \), we first need to rearrange the equation to standard form:

\[
x^2 - 16x - 13 - 4 = 0
\]

This simplifies to:

\[
x^2 - 16x - 17 = 0
\]

Next, we can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -16 \), and \( c = -17 \).

First, we calculate the discriminant:

\[
b^2 - 4ac = (-16)^2 - 4(1)(-17) = 256 + 68 = 324
\]

Now, we can apply the quadratic formula:

\[
x = \frac{-(-16) \pm \sqrt{324}}{2(1)} = \frac{16 \pm 18}{2}
\]

This gives us two solutions:

1. \( x = \frac{16 + 18}{2} = \frac{34}{2} = 17 \)
2. \( x = \frac{16 - 18}{2} = \frac{-2}{2} = -1 \)

Thus, the zeros of the equation are:

\[
x = -1, \quad x = 17
\]

So, in order of size:

\[
x = -1, \quad x = 17
\]