Question

What are the values of x^2 - 16x - 13 = 4? Enter the lesser value first

Answer 1

To solve the equation \( x^2 - 16x - 13 = 4 \), we first move all terms to one side of the equation:

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

Simplifying this gives:

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

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

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

For our equation \( ax^2 + bx + c = 0 \):

- \( a = 1 \)
- \( b = -16 \)
- \( c = -17 \)

Now, we substitute these values into the quadratic formula:

1. Calculate the discriminant:

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

2. Then, we calculate \( x \):

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

Now, we calculate the two potential solutions:

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

Thus, the solutions to the equation \( x^2 - 16x - 13 = 4 \) are \( x = -1 \) and \( x = 17 \).

The lesser value is \( -1 \) and the greater value is \( 17 \).

The final answer is:

\[
-1, 17
\]