Question

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


Enter the smaller of the 2 values first.

Answer 1

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

1. Subtract 4 from both sides:

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

which simplifies to:

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

Next, we can use the quadratic formula to find the values of \( x \):

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

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

Now we calculate \( b^2 - 4ac \):

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

Thus,

\[
b^2 - 4ac = 256 + 68 = 324.
\]

Now substituting back into the quadratic formula:

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

This gives us two potential solutions:

1.
\[
x = \frac{16 + 18}{2} = \frac{34}{2} = 17,
\]

2.
\[
x = \frac{16 - 18}{2} = \frac{-2}{2} = -1.
\]

Therefore, the zeros of the equation are \( -1 \) and \( 17 \).

Since we are asked to enter the smaller value first, the final answer is:

\[
\boxed{-1}, 17.
\]