Question

Find the zeros of the quadratic equation: x2−16x−13=4


Enter the smaller of the 2 values first.

(1 point)
x=

, x=

Answer 1

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

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

This simplifies to:

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

Now we will use the quadratic formula, which is given by:

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

Here, \( a = 1 \), \( b = -16 \), and \( c = -17 \).

Now we calculate the discriminant:

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

Next, we find the square root of the discriminant:

\[
\sqrt{324} = 18
\]

Now we can apply the quadratic formula:

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

Calculating the two possible values for \( x \):

1. \( x_1 = \frac{16 + 18}{2} = \frac{34}{2} = 17 \)
2. \( x_2 = \frac{16 - 18}{2} = \frac{-2}{2} = -1 \)

Thus, the solutions are \( x = -1 \) and \( x = 17 \).

To enter the smaller value first:

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