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 move all terms to one side of the equation:

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

This simplifies to:

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

Next, we can use the quadratic formula to find the zeros of the equation, which states that for any quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) can be found using:

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

In our equation, \( a = 1 \), \( b = -16 \), and \( c = -17 \).

Calculating the discriminant:

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

Now we apply the quadratic formula:

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

Calculating the two possible values:

1. When we take the positive root:
\[
x = \frac{{16 + 18}}{2} = \frac{34}{2} = 17
\]

2. When we take the negative root:
\[
x = \frac{{16 - 18}}{2} = \frac{-2}{2} = -1
\]

Thus, the zeros of the quadratic equation \( x^2 - 16x - 17 = 0 \) are \( -1 \) and \( 17 \).

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

\[
-1, 17
\]