1. 1.5 Applications and Modeling with Quadratic Equations
Geometry Problems
Using the Pythagorean Theorem
Height of a Projected Object
Modeling with Quadratic Equations

2. SOLVING A PROBLEM INVOLVING VOLUME
Example 1
A piece of machinery produces rectangular sheets of metal such that the length is three times the width. Equal-sized squares measuring 5 in. on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in.3, find the dimensions of the original piece of metal.

3. SOLVING A PROBLEM INVOLVING VOLUME
Example 1
Solution
Step 1 Read the problem. We must find the dimensions of the original piece of metal.

4. Solution SOLVING A PROBLEM INVOLVING VOLUME Example 1
Step 2 Assign a variable. We know that the length is three times the width. Let x = the width (in inches) and thus, 3x = the length.
The box is formed by cutting = 10 in. from both the length and the width.
3x – 10 5
x – 10 x 3x
x – 10
3x – 10 5

5. SOLVING A PROBLEM INVOLVING VOLUME
Example 1
Solution
Step 3 Write an equation. The formula for the volume of a box is V = lwh.
Volume
length 
width 
height =
(Note that the dimensions of the box must be positive numbers, so 3x – 10 and x – 10 must be
greater than 0, which implies
and
These are both satisfied when x > 10.)

6. Step 4 Solve the equation.
SOLVING A PROBLEM INVOLVING VOLUME
Example 1
Solution
Step 4 Solve the equation.
Multiply.
Subtract 1435 from each side.
Divide each side by 5.
Factor.

7. The width cannot be negative.
SOLVING A PROBLEM INVOLVING VOLUME
Example 1
Solution
Step 4 Solve the equation. or
Zero-factor property or
Solve each equation.
The width cannot be negative.

8. SOLVING A PROBLEM INVOLVING VOLUME
Example 1
Solution
Step 5 State the answer. Only 17 satisfies the restriction x > 10. Thus, the dimensions of the original piece should be 17 in. by 3(17) = 51 in.

9. SOLVING A PROBLEM INVOLVING VOLUME
Example 1
Solution
Step 6 Check. The length of the bottom of the box is 51 – 2(5) = 41 in. The width is
17 – 2(5) = 7 in. The height is 5 in. (the amount cut on each corner), so the volume of the box is

10. Using the Pythagorean Theorem
Example 2 requires the use of the Pythagorean theorem for right triangles. Recall that the legs of a right triangle form the right angle, and the hypotenuse is the side opposite the right angle.
Hypotenuse c
Leg a
Leg b

11. Pythagorean Theorem
In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
Hypotenuse c
Leg a
Leg b

12. APPLYING THE PYTHAGOREAN THEOREM
Example 2
A piece of property has the shape of a right triangle. The longer leg is 20 m longer than twice the length of the shorter leg. The hypotenuse is 10 m longer than the length of the longer leg. Find the lengths of the sides of the triangular lot.

13. Step 1 Read the problem. We must find the lengths of the three sides.
APPLYING THE PYTHAGOREAN THEOREM
Example 2
Solution
Step 1 Read the problem. We must find the lengths of the three sides.
Step 2 Assign a variable.
Let x = the length of the shorter leg (in meters).
Then 2x + 20 = the length of the longer leg, and
(2x + 20) + 10 or 2x + 30 = the length of the hypotenuse.
2x + 30 x
2x + 20

14. Solution Step 3 Write an equation. APPLYING THE PYTHAGOREAN THEOREM
Example 2
Solution
Step 3 Write an equation.
The hypotenuse is c.
Substitute into the Pythagorean theorem.

15. Remember the middle term here. Remember the middle term here.
APPLYING THE PYTHAGOREAN THEOREM
Example 2
Solution
Step 4 Solve the equation.
Remember the middle term here.
Remember the middle term here.
Square the binomials.

16. Step 4 Solve the equation.
APPLYING THE PYTHAGOREAN THEOREM
Example 2
Solution
Step 4 Solve the equation.
Standard form
Factor. or
Zero-factor property or
Solve each equation.

17. APPLYING THE PYTHAGOREAN THEOREM
Example 2
Solution
Step 5 State the answer. Since x represents a length, – 10 is not reasonable. The lengths of the sides of the triangular lot are 50 m, 2(50) + 20 = 120 m, and 2(50) + 30 = 130 m.
Step 6 Check. The lengths 50, 120, and 130 satisfy the words of the problem, and also satisfy the Pythagorean theorem.

18. Height of a Projected Object
If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height of s0 feet, with initial velocity v0 feet per second is given by the equation
Here t represents the number of seconds after the object is projected. The coefficient of t 2, – 16, is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon and other planets.

19. SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT
Example 3
If a projectile is launched vertically upward from the ground with an initial velocity of 100 ft per sec, neglecting air resistance, its height s (in feet) above the ground t seconds after projection is given by

20. (a) After how many seconds will it be 50 ft above the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT
Example 3
(a) After how many seconds will it be 50 ft above the ground?
Solution
We must find value(s) of t so that height s is 50 ft.
Let s = 50 in the given equation.
Standard form
Divide by −2.

21. (a) After how many seconds will it be 50 ft above the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT
Example 3
(a) After how many seconds will it be 50 ft above the ground?
Solution
Quadratic formula
Simplify. or
Use a calculator.

22. (a) After how many seconds will it be 50 ft above the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT
Example 3
(a) After how many seconds will it be 50 ft above the ground?
Solution or
Both solutions are acceptable, since the projectile reaches 50 ft twice—once on its way up (after 0.55 sec) and once on its way down (after 5.70 sec).

23. (b) How long will it take for the projectile to return to the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT
Example 3
(b) How long will it take for the projectile to
return to the ground?
Solution When the projectile returns to the ground, the height s will be 0 ft.
Let s = 0.
Factor. or
Zero-factor property or
Solve each equation.

24. (b) How long will it take for the projectile to return to the ground?
SOLVING A PROBLEM INVOLVING PROJECTILE HEIGHT
Example 3
(b) How long will it take for the projectile to
return to the ground?
Solution
The first solution, 0, represents the time at which the projectile was on the ground prior to being launched, so it does not answer the question. The projectile will return to the ground 6.25 sec after it is launched.

25. ANALYZING TROLLEY RIDERSHIP
Example 4
The I-Ride Trolley service carries passengers along the International Drive Resort Area of Orlando, Florida. The bar graph shows I-Ride Trolley ridership in millions. The quadratic equation
models ridership from 2000 to 2010, where y represents ridership in millions and x = 0 represents 2000, x = 1 represents 2001, and so on.

26. ANALYZING TROLLEY RIDERSHIP
Example 4
(a) Use the model to determine ridership in Compare the result to the actual ridership figure of 2.1 million.
Solution
Since x = 0 represents the year 2000, use x = 8 to represent 2008.
Given model
Let x = 8.
Use a calculator.
The prediction is about 0.1 million (that is, 100,000) less than the actual figure of 2.1 million.

27. ANALYZING TROLLEY RIDERSHIP
Example 4
(b) According to the model, in what year did ridership reach 1.9 million?
Solution
Let y = 1.9 in the model.
Standard form
Quadratic formula or
Use a calculator.

28. ANALYZING TROLLEY RIDERSHIP
Example 4
(b) According to the model, in what year did ridership reach 1.9 million?
Solution
The year 2003 corresponds to x = 3.3. Thus, according to the model, ridership reached 1.9 million in the year This outcome closely matches the bar graph and seems reasonable. The year 2011 corresponds to x = The model predicts that ridership will be 1.9 million again in the year 2011.