1. Chapter 2: Equations and Inequalities 2.3: Applications of Equations
Essential Question: What are the different methods to solve a quadratic equation?

2. 2.3 Applications of Equations
Guidelines for solving applied problems
Read the problem carefully, and determine what is asked for.
Label the unknown quantities with variables
Draw a picture of the situation, if appropriate
Translate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language.
Consolidate the mathematical information into an equation in one variable that can be solved or an equation in two variables that can be graphed.
Solve for at least one of the unknown quantities.
Find all remaining unknown quantities by using the relationships given in the problem.
Check and interpret all quantities found in the original problem.

3. 2.3 Applications of Equations
Example 1: Number Relations
The average of two real numbers is , and their product is Find the two numbers.
Solution
Two equations:
Solve one equation for one variable, and then substitute.

4. 2.3 Applications of Equations
Example 1 (Continued)
a a+1683=0
Option 1 → Graph
Answers are where the graph crosses the x-axis
Option 2 → Ye old Quadratic Equation

5. 2.3 Applications of Equations
Example 1 (Finishing)
Check the answers
If a = 44
Plug into the ab=1683 equation to find that b=38.25
If a = 38.25
Plug into the same function to find that b=44
Check
The average of 44 & is
The two numbers are 44 and 38.25

6. 2.3 Applications of Equations
Example 2: Dimensions of a Rectangle
A rectangle is twice as wide as it is high. If it has an area of 24.5 square inches, what are its dimensions?
Two equations
Substitute and solve

7. 2.3 Applications of Equations
Example 2 (Continued)
It’s not possible to have a negative height, so the only value we have to check is h=3.5 inches
If the width is twice the height, then the width = 2(3.5) = 7 inches
Check your answer: (3.5 in)(7 in) = 24.5 in2
The width is 7 inches and the height is 3.5 inches

8. 2.3 Applications of Equations
Example 4: Interest Applications
I = Prt
I = Interest
P = Principal (initial invested amount)
r = rate (percentage written as a decimal)
t = time (in years)
A high-risk stock pays dividends at a rate of 12% per year, and a savings account pays 6% interest per year. How much of a $9000 investment should be put in the stock and how much should be put in savings to obtain a return of 8% per year on the total investment?

9. 2.3 Application of Equations
Example 4: Interest Applications (continued)
Let s be the amount invested in stock. The rest of the money ($9000 – s) is the amount invested in savings
Translate English into math:
Individual interest added together = total interest earned
(stock at 12%) + (savings at 6%) = 8% of $9000
0.12s (9000 – s) = 0.08(9000) [Distribute]
0.12s – 0.06s = 720 [Combine like terms]
0.06s =
0.06s ÷ 0.06 = 180 ÷ 0.06
s = 3000

10. Application of Equations
Example 5: Distance Applications
d = rt
d = distance
r = rate
t = time
You can convert the equation if necessary
r = d/t
t = d/r
A pilot wants to make an 840-mile round trip from Cleveland to Peoria and back in 5 hours flying time. There will be a headwind of 30mph going to Peoria, and it is estimated that there will be a 40mph tailwind returning to Cleveland. At what constant engine speed should the plane be flown?

11. 2.3 Application of Equations
Example 5: Distance Applications (continued)
Let r be the engine speed of the plane
Headwind slows the velocity by 30 mph
Tailwind increases the velocity by 40 mph
Cleveland to Peoria
Distance = 420
Actual velocity = r – 30
Time = 420/(r – 30)
Peoria to Cleveland
Actual velocity = r + 40
Time = 420/(r + 40)

12. Application of Equations
Example 5: Distance Applications (continued)
The total time is going to be 5 hours, so

13. 2.3 Application of Equations
Example 5: Distance Applications (concluded)
r r = 0
This can be factored (you can use the quadratic equation as well of course)
(r – 170)(r + 12) = 0
r – 170 = 0 or r + 12 = 0
r = 170 or r = -12
Because you can’t fly at a negative rate, the plane must fly at a constant rate of 170 mph

14. 2.3 Application of Equations
Example 8: Mixture Problem
A car radiator contains 12 quarts of fluid, 20% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the resulting mixture is 50% antifreeze?
Let x be the number of quarts of fluid to be replaced by pure antifreeze.
When x quarts are drained, there are 12 – x quarts of fluid left in the radiator, 20% of which is antifreeze.

15. 2.3 Application of Equations
Example 8: Mixture Problem (continued)
Translate English into math
20% of (12 – x) + x = 50% of 12
0.2(12 – x) + x = 0.5(12) [Distribute]
2.4 – 0.2x + x = 6 [Combine like terms]
0.8x – 2.4 = 6 – 2.4
0.8x ÷ 0.8 = 3.6 ÷ 0.8
x = 4.5
4.5 quarts should be drained and replaced with pure antifreeze

16. 2.3 Applications of Equations
Assignment
Pages
Wednesday: 9, 15, 17, 25
Thursday: 11, 13, 19, 21, 23
For #25, you’re going to want to solve it by graphing and finding the x-intercept(s). Make sure you alter your window so that you can see a solution.