Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Activity

In a previous lesson, you saw an equation that defines the height of a potato as a function of time after it was launched from a mechanical device. Here is a different function modeling the height of a potato, in feet, $t$ seconds after being fired from a different device:

$f (t) = - 16 t^{2} + 80 t + 64$

1.

What equation would we solve to find the time at which the potato hits the ground?

Enter an equation.

Compare your answer:

$- 16 t^{2} + 80 t + 64 = 0$

Since the potato hits the ground at a height of 0 feet, set the equation equal to 0.

2.

Use any method except graphing to find a solution to the equation from the previous question.

Enter a solution to the equation.

Compare your answer:

An approximate solution is about 5.7 seconds by using trial and error.

Building Character: Curiosity

An illustrated person kneels outdoors, holding a magnifying glass and examining orange flowers, with a question mark in a speech bubble above their head. Pine trees and clouds are in the background.

Curiosity is the desire to learn and understand new things. When you are curious about something, you process it on a deeper level than just basic understanding.

Think about your current sense of curiosity. Are the following statements true for you??

I got so absorbed in learning that I lost track of time.
I took the initiative to learn more about one of my interests.

Don’t worry if none of these statements are true for you. Developing this trait takes time. Your first step starts today!

Self Check

After a flashlight is turned on, it gets brighter as it warms up, and then it grows dimmer as the battery dies. The equation modeling the brightness it emits, in lumens, $h$ hours after being turned on is shown here.

$f(h) = -8h^2 + 30h + 50$

Which equation would you solve to determine when the flashlight dies and stops emitting lumens?

$-8h^2 + 30h + 50 = 0$
$-8h^2 + 30h = 50$
$8h^2 + 30h + 50 = 0$
$-8h^2 + 30h - 50 = 0$

Additional Resources

Using Various Methods to Solve Quadratic Equations

A tennis ball is thrown out of the window of a building. The equation modeling the height of the tennis ball, in feet, $t$ seconds after being thrown is shown here.

$f (t) = - 24 t^{2} + 33 t + 30$

Let’s find the equation we would solve to find the time at which the ball hits the ground.

When the ball hits the ground, its height will be 0. So, we set the equation to 0.

$- 24 t^{2} + 33 t + 30 = 0$

The equation above represents the ball when it hits the ground. The solution to this equation represents the time it takes from when the ball is thrown for it to hit the ground.

Let’s try to solve the equation.

Step 1 - Write the equation to be solved.

$- 24 t^{2} + 33 t + 30 = 0$

Step 2 - Subtract 30 from each side of the equation.

$- 24 t^{2} + 33 t = - 30$

Step 3 - Divide each side of the equation by 3.

$- 8 t^{2} + 11 t = - 10$

Let’s try to solve the remaining equation using guess and check.

When $t = 3$ , the expression $- 8 t^{2} + 11 t$ has a value of $- 8 (3)^{2} + 11 (3)$ , or $- 39$ . This value is not equal to the $- 10$ we needed to find.

When $t = 2$ , the expression $- 8 t^{2} + 11 t$ has a value of $- 8 (2)^{2} + 11 (2)$ , or $- 10$ . This value is equal to the $- 10$ we needed to find.

The equation is true when $t = 2$ , so the ball took 2 seconds to hit the ground. This method works but can be more difficult when the solution is a decimal. Graphing can also work, but it is time consuming and results in an approximate solution.

Try it

Try It: Using Various Methods to Solve Quadratic Equations

An arc of water flows from a water fountain down to a pool below it. The height of the arc above the pool, in inches, can be modeled using the equation $f (x) = - 2 x^{2} + 5 x + 3$ , where $x$ is the horizontal distance the water has traveled, in inches, from the water fountain.

1. Which equation would you solve to determine how far the arc of water has traveled horizontally when it reaches the pool below?

2. Solve the equation using a non-graphing method.

Here is how to write and solve the equation representing the arc of water from the water fountain:

The equation $- 2 x^{2} + 5 x + 3 = 0$ represents the arc of the water when it reaches the pool below.
To solve the equation $- 2 x^{2} + 5 x + 3 = 0$ , let’s try using guess and check with substitution.

First, we try $x = 2$ .

$- 2 (2)^{2} + 5 (2) + 3 \overset{?}{=} 0$

$5 \neq 0$

Now, let’s try $x = 3$ .

$- 2 (3)^{2} + 5 (3) + 3 \overset{?}{=} 0$

$0 = 0$

Since the equation is true, $x = 3$ is correct.

The arc of water has traveled 3 inches horizontally when it reaches the pool below the fountain.

8.2.2 Exploring Non-Graphing Strategies to Solve Quadratic Equations

Activity

Additional Resources

Using Various Methods to Solve Quadratic Equations

Try It: Using Various Methods to Solve Quadratic Equations