Use the quadratic equation x2−18x=−80

Question

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: \( 10 \)