Question

Use the quadratic equation x2−18x=−80
to answer the questions.

Part A: Suppose the equivalent quadratic equation is written x2−18x+c=−80+c
. What value of c makes the equation a perfect square trinomial?

Part B: Factor the perfect square trinomial, and rewrite the equation as (x−h)2=k

Part C: Solve the quadratic equation and find a solution to x. Enter the lesser number.

Part D: Solve the quadratic equation and find a solution to x. Enter the greater number.

(4 points)
Part A:

Part B:

Part C:

Part D:

Answer 1

To solve the problem step-by-step, let's go through each part.

### Part A:
We start with the equation \( x^2 - 18x = -80 \). We want to rewrite it in the form \( x^2 - 18x + c = -80 + c \) so that the left side is a perfect square trinomial.

A perfect square trinomial can be formed using the formula:
\[
(x - a)^2 = x^2 - 2ax + a^2
\]
For our equation, \( -2a = -18 \) implies that \( a = 9 \) (since \( 2a = 18 \)).

To make this a perfect square trinomial, we need to add \( a^2 \) to both sides:
\[
c = 9^2 = 81
\]

Thus, we replace \( c \) in our equation:
\[
x^2 - 18x + 81 = -80 + 81
\]

So, \( c \) is:

**Part A:** \( c = 81 \)

### Part B:
Now we rewrite the equation as:
\[
x^2 - 18x + 81 = 1
\]
The left-hand side \( x^2 - 18x + 81 \) can be factored as:
\[
(x - 9)^2 = 1
\]

So, we have:
**Part B:** \( (x - 9)^2 = 1 \)

### Part C:
Next, we solve for \( x \):
Taking the square root of both sides gives:
\[
x - 9 = \pm 1
\]
This results in two equations:
1. \( x - 9 = 1 \) ⇒ \( x = 10 \)
2. \( x - 9 = -1 \) ⇒ \( x = 8 \)

We need the lesser number:
**Part C:** \( x = 8 \)

### Part D:
Now, from the previous results, we find the greater number, which is:
**Part D:** \( x = 10 \)

### Summary of Answers:
- **Part A:** \( 81 \)
- **Part B:** \( (x - 9)^2 = 1 \)
- **Part C:** \( 8 \)
- **Part D:** \( 10 \)