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 rearrange it into the standard form \( ax^2 + bx + c = 0 \).

1. Subtract 4 from both sides:

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

Now we can use the quadratic formula to solve for \( x \). The quadratic formula is given by:

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

In our equation, the coefficients are:
- \( a = 1 \)
- \( b = -16 \)
- \( c = -17 \)

2. Next, we calculate the discriminant \( b^2 - 4ac \):

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

3. Now we'll use the quadratic formula to find the values of \( x \):

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

4. Now we calculate the two solutions:

- For the positive case:

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

- For the negative case:

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

Thus, the solutions (or zeros) of the equation are \( -1 \) and \( 17 \).

The smaller of the two values is:

\[
\boxed{-1}
\]

And the larger value is \( 17 \).