CA-250 Chapter 9. Algebra
SECTION 9.7 CLASS ACTIVITY 9-BB
Doing Rocket Science by Reasoning about the Structure of Quadratic Equations
CCSS CCSS SMP7
1. Rocket scientists determined that $t$ seconds after launch, a rocket's height, $h$, in feet above the ground will be given by the equation
\[
h=16(2+t)(22-t)
\]
a. Describe the structure of the expression on the right of the equal sign.
b. Reason about the structure of $16(2+t)(22-t)$ to determine which values for $t$ will make the expression have a value of 0 . To help your thinking: when you multiply numbers, how can the result be 0 ?

Step 1 ：The expression on the right of the equal sign is a quadratic equation in the form of \(ax(bx+c)(dx-e)\) where a, b, c, d, and e are constants and x is the variable. In this case, a = 16, b = 1, c = 2, d = -1, and e = 22.

Step 2 ：The values for t that will make the expression have a value of 0 are the roots of the equation. In a multiplication operation, the result can be 0 if any of the factors is 0. Therefore, we need to find the values of t that make either \((2+t)\) or \((22-t)\) equal to 0.

Step 3 ：The roots of the equation are -2 and 22. These are the values for t that will make the expression \(16(2+t)(22-t)\) equal to 0.

Step 4 ：Final Answer: The values for t that will make the expression \(16(2+t)(22-t)\) have a value of 0 are \(\boxed{-2}\) and \(\boxed{22}\).