Essential Question: What are the different methods to solve a quadratic equation?

Guidelines for solving applied problems1. Read the problem carefully, and determine what is

asked for.2. Label the unknown quantities with variables3. Draw a picture of the situation, if appropriate4. Translate the verbal statements in the problem and the

relationships between the known and unknown quantities into mathematical language.

5. Consolidate the mathematical information into an equation in one variable that can be solved or an equation in two variables that can be graphed.

6. Solve for at least one of the unknown quantities.7. Find all remaining unknown quantities by using the

relationships given in the problem.8. Check and interpret all quantities found in the original

problem.

Example 1: Number Relations◦ The average of two real numbers is 41.125, and

their product is 1683. Find the two numbers.◦ Solution

Two equations:

Solve one equation for one variable, and then substitute.

41.1252

1683

a b

ab

41.1252

82.25

82.2

2

5

2a b

a a ab

b a

2 2 2

2

(82.25 )

82

1683

1683

82.25 1683.25 82

0

.25

82.25 1683

a

a a a

ab

a

a a

a a

a

Example 1 (Continued)◦ a2-82.25a+1683=0◦ Option 1 → Graph

Answers are where the graph crosses the x-axis◦ Option 2 → Ye old Quadratic Equation

2( 82.25) ( 82.25) 4(1)(1683)

2(1)

82.25 6765.0625 6732

2

82.25 33.0625

2

82.25 5.75

288 76.5

or 2 2

44 or 38.25

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 & 38.25 is 41.125

◦ The two numbers are 44 and 38.25

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

2

24.5

w h

wh

2

2

24.5

2 24.5

12.25

(2 )

2 2

h

h

h

h

12.25

3.5

h

h

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

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?

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 + 0.06(9000 – s) = 0.08(9000) [Distribute] 0.12s + 540 – 0.06s = 720 [Combine like terms] 0.06s + 540 - 540 = 720 - 540 0.06s ÷ 0.06 = 180 ÷ 0.06 s = 3000

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?

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 Distance = 420 Actual velocity = r + 40 Time = 420/(r + 40)

Example 5: Distance Applications (continued)◦ The total time is going to be 5 hours, so

420 4205

30 40r r

( 30)420 420

530

( 34

0 ( 300

) )r r

r r r

420( 30)420 5( 30)

( 4( 40) ( 40)

0( 40)

)r r

rr

rr

420( 40) 420( 30) 5( 30)( 40)r r r r

2420 16800 420 12600 5( 10 1200)r r r r 2840 4200 5( 10 1200)r r r

2168 840 10 1200r r r 20 158 2040r r

Example 5: Distance Applications (concluded)◦ r2 - 158r - 2040 = 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

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.

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 – 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

Amount of antifreeze Amount of x quarts of

in radiator after draining antifreeze inantifreeze

x quarts of fluid final mixture

Assignment◦ Pages 105-106◦ 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.