Question

find the zeros of the quadratic equation: x^2 - 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 rewrite the equation in standard form by moving all terms to one side:

\[
x^2 - 16x - 13 - 4 = 0
\]
\[
x^2 - 16x - 17 = 0
\]

Next, we can use the quadratic formula to find the values of \( x \). The quadratic formula is given by:

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

In our equation \( a = 1 \), \( b = -16 \), and \( c = -17 \). We can now calculate the discriminant (\( b^2 - 4ac \)):

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

Now we substitute the values into the quadratic formula:

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

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 quadratic equation are \( -1 \) and \( 17 \).

Since we need to enter the smaller value first, the answer is:

\[
-1, 17
\]