INTRODUCTION

We are here to make your life easier when it comes to quadratic word problems. First things first… try not to get fustrated, if it seems like it is getting hard you are quite right, but we are going to make it so it’s a piece of cake. If you are getting a little upset take a break, relax and watch some t.v, but get back at it!

First we will explain the quadratic equation and give you some practice problems. This is to verify your knowledge and ability to work with equations, before you try harder word problems (not for long).

Ok, next you will learn how to analyze word problems, by arranging information that is given to you, to make it less complicated. Once you have arranged the information you will be able to use your previous knowledge on quadratic equations to solve the word problems!

At the very end you will finally be able to put yourself to a test and see how much you have learned by writing the test.

GOOD LUCK AND DON’T GIVE UP !!!!!

THE QUADRIATIC EQUATION

A quadratic function is the name for a function where the highest order of X (the variable) is 2. This means that X is squared. The generalized equation for a quadratic function that has 2 as the highest power is:

F(x)= ax2 + bx + c

The letters in this equation are constants, which represent and hold the place of numbers, which will appear in actual equations.

Example f(x)= 5x2 + 24x + 8 A = 5

B = 24 C = 8

The roots, zeros, or x –intercepts of a function are the points where f(x)=0. In other words you fill zero in for y and solve for x. The quadratic equation can be rearranged to solve for x. After it has been rearranged it is known as the quadratic formula. The quadratic formula looks like this…

_____X = -b +or - b 2 -4ac

2a

This equation will always give you two different answers, one using the + sign and one using the – sign.

The numbers go into the formula according to the letters just as they did with the equation. Example f(x)= 5x2 + 24x + 8 __________

A = 5 x = -24+ or - 24²-4(5)(8) B = 24 2(5)

C = 8Now simply solve the equation for x. Don’t forget you will have 2 values for x.

_________ X= -24+ or - 24²-4(5)(8)

2(5) _______

X= -24+or- 576-160 10

___ ___ X=-24- 416 X=-24+ 416

10 10

X=-24-20.4 X=-24+20.410 10

X=-44.4 X= -3.610 10

X=-4.44 X=-0.36

You have just seen the quadratic formula solve for x by simply plugging the numbers into the formula from the equation. Lets do one more for practice.Example 2

2.1x²-3.2x =5.2

First we have to put this equation back into the format of the quadratic equation:

2.1x²-3.2x-5.2 = 0

Now we have to plug our quadratic equation into the quadratic formula: _________________

X = - (-3.2) + or- √ (-3.2)² - 4 (2.1) (-5.2) 2 (2.1)

Now simply do the math and solve the equation! _____________

X = 3.2 + or - √ 10.24 – (-43.68) 4.2

____ ____X = 3.2 - √53.92 X = 3.2 + √53.92

4.2 4.2

X = -0.99 X = 2.51

Now we have successfully solved for x! Now, if you think you’re ready to try some yourself there are some practice problems just below. If you still are not completely sure of your self you may want to read over the above once more before giving them a try!

PRACTICE PROBLEMS(solving for x using the quadratic formula)

Instructions: Below are some examples of equations. You will have to rearrange the equations (if necessary) and then solve for x using the quadratic formula. An answer key is provided at the back of this tutorial kit. Only look at the answer key when you have COMPLETELY FINISHED the problems. Good Luck and don’t give up!

1. 3x²-5x-1=0

2. 15x²+26x+12=0

3. 6x²-27x+20=0

4. x²-2x+5=0

5. 4x²-24=10x

6. 6x²-5x=4

7. 9-6x=-x²

8. 2(x+4)=x²

9. (2x+1)²-(2x+1)=x²-x

10.(1-x) (x+4) + 2 (2x+1)² =6 3

10. 3x=2+4 x

Hopefully you have successfully completed the practice problems and verified your answers, if not you may want to go give them another try. It is important that you are capable of doing these standard problems before trying to take things a step further…Word Problems! If you have successfully finished these problems then lets continue on!

HOW TO BREAK DOWN WORD PROBLEMS!

Solving word problems is a skill that can be learned but has to be practiced. Here are some helpful hints to remember when trying to solve them to make your life easier!

• Read the problem carefully until you understand what information is given and what information is asked. You should be able you answer the following questions:

a) What information do we know? b) What information are we asked to find? c) Is there a need to introduce any variables? If so what are they?

• Write down what information that is given and the information that is asked. Determine if unnecessary information is given.

•Translate the problem into mathematics by writing the appropriate equation. (In our case more than likely the quadratic equation)

•Solve the equation!

•Write down the answer with the appropriate unit and any other information needed in a proper sentence. (Ex. The speed of the car is 75 miles/hour)

•Check you answer in the original word problem and verify that it makes sense. “Mary is 135 years old” is clearly wrong.

• If your answer makes sense give yourself a tap on the back!

Now that you know how to correctly analyze word problems lets go through a couple of examples!

WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

So far all we have done is plug in numbers from one equation to another. Now we are going to take it one step farther, not only are we going to plug in the numbers, but we have to find the equations as well. We will go through this lesson the same way as the last one: we will do two examples together than there will be questions to do on your own. Remember to use the steps above and try not to get frustrated! Here we go…

The area of a rectangular window is 360 square centimeters. If the length of the window is bigger then the width of the window by 1 centimeter, what are the dimensions?(Area formula is equal to Length • width)

Lets take it one step at a time:•Read the problem carefully and answer the following questions:

a) What information do we know?b) What information are we asked to find?c) Is there a need to introduce variables? If so what are they?

• Write down the information given and the information that is asked:

a) What information we know: Area=360 square centimeters Length= Width + 1 centimeter

b) What information we are asked to find: The dimensions of the windowc) Variable: Width (since we know the least about it) so we can state: L = W + 1

• Translate the problem into mathematics and write the appropriate equation.

Because we know that Area is equal to Length time Width our equation would be:

360 = W (W + 1)We can now take this formula and put it into the quadratic equation so it looks like this:

0 = W² + W – 360• Solve the equation!

-We need two numbers that multiply to - 360 and that add to one. -Because there are not two numbers that multiply to 360 and add to 1 we

will need to use the quadratic formula:

We know that: A =1 B= 1 C= -360 Now we plug these numbers into the equation:

____________X= -1 +or- (1)²-4 (1) (-360)

2 ______

X= -1+or-√1+1140 2

W = 18.48 W = - 19.48

• Write down the answer with the appropriate unit and any other information needed in a proper sentence:

The width of the window is 18.48 centimeters or –19.48 centimeters.

• Check you answer in the original word problem and verify that it makes sense:Notice that since we are working with dimensions they cannot be negative! Therefore the width has to be 18.48 centimeters. Now we can figure out the length because we know that Length is equal to Width +1 centimeter. Therefore:

W + 1 = 18.48 centimeters + 1 = 19.48 centimeters = length

Answer: The dimensions of the window are: Width = 18.48 centimeters Length = 19.48 centimeters*To verify that our answer is correct we can now plug our answer into the area equation:

Area = Length • WidthArea = 19.48cm • 18.48cmArea = 360cm²

• If your answer makes sense give yourself a pat on the back!!WELL DONE!

Lets try another example since that one was so easy…

A radiation control point is set up near a solid waste disposal facility. The pad on which the facility is set up measures 20 feet by 30 feet. If the health physicist sets up a controlled walkway around the pad that reduces the area by 264 square feet, how wide is the walkway?

• Read the problem carefully and answer the following questions: a) What information do we know?

b) What information are we asked to find?c) Is there a need to introduce variables? If so what are they?

• Write down the information given and the information that is asked:

a) What information we know: Pad measures 20 feet by 30 feetWalkway that reduces the pad area by 264 square feet

b) What information we are asked to find: The width of the walkwayc) Variable: Width of the walkway

• Translate the problem into mathematics and write the appropriate equation.30 - 2x = Length of Reduced Pad 20 - 2x = Width of Reduced PadArea of Reduced Pad = (Length of Reduced Pad)(Width of Reduced

Pad)With this information we are able to come up with our quadratic equation:

600 – 264 = (30 – 2x)(20 – 2x)336 = 600 – 100x + 4x2

•Solve the equation!

4x2 - 100x + 264 = 0We know that: A=4

B=100 C=264

Now we can substitute these numbers into the quadratic formula: _______________

X= -(-100) +or-√(-100)²-4 (4) (264)2(4)

___________X= 100 +or-√10 000 – 4 224

8

X=22X=3

• Write down the answer with the appropriate unit and any other information needed in a proper sentence: The width of the walkway is 22 feet or 3 feet.

• Check you answer in the original word problem and verify that it makes sense:The two roots are x = 22 feet and x = 3 feet. Since x = 22 feet is not physically meaningful, the answer is only x = 3 feet.

Answer: The width of the walkway is 3 feet wide.

We can also check to see if our answer is correct in this problem as well by using the formula for area:The area of the reduced area pad is 264 square feet less than the area of the original pad therefore 600 – 264 = (20 – 2x)(30-2x)

336 = [20 – 2(3)][30 – 2(3)] 336 = (20 – 6)(30 – 6) 336 = (14)(24) 336 = 336

As you can see our answer was correct! Well Done!

Now that we have done a couple of word problems together why not try some yourself? Just remember take your time and take it one step at a time. Good Luck!

PRACTICE WORD PROBLEMS!(using the quadratic formula)

Instructions: Using everything that we have gone over up to this point you are now ready to try and solve some word problems yourself! Be sure to read over them carefully and use the steps! When you have completed the questions you can check your answers at the end.

1. An outdoor decorator was hired to design canopies for the outside of shops all the way around the block. The trouble was the decorator did not know how far out the canopy should stretch, if it was too short the rain water would run off and soak the people on the sidewalk. He had to decide how far out was long enough to pass over peoples heads but not so long as to be a waste of materials. The decorator did know the buildings dimensions (the shops were all attached) were 60m by 40m and the area of the sidewalk was 2815m². How far out do the canopies extend?

2. An apartment building 27m by 50m is built on a lot that is 383m². If a parking strip surrounds the apartment building, how wide is the parking strip?

3. A community pool was under construction to make it bigger. It was 15m by 8m and in the construction the area doubled by adding the same amount of length onto one end and one side. How much was added?

4. A group of friends decide to go to a movie. The tickets cost $50 altogether. It is one person’s birthday so the others split her ticket cost which raises theirs $1.50. How many friends are going to the movie?

5. If a dinner out at a restaurant cost $600 but one member present cannot pay, and everyone else’s cost goes up by only $2.50. How many people are present?

6. A king decided he needed a moat for around his castle. He only wanted the moat to extend far enough so the area of the moat and the land the castle was on was 16000m2. The castle was 150m by 75m. How far out from the castle walls will the moat extend?

7. How much must be added to the shorter side of a rectangle 8cm long and 6cm wide in order to form a new rectangle that’s diagonal is 7cm longer than the diagonal of the original rectangle.

8. Artillerymen on a hillside are trying to hit a target behind a mountain on the other side of the river. Their cannon is at (x,y) = (3,250), where x is in kilometers and y is in meters. The target is at (x,y) = (-2,50). In order to avoid hitting the mountain on the other side of the river, the projectile from the cannon must go through the point (x,y) = (-1,410). If a= -80, b=120 and c=610 :

a) Write the particular equation of the parabolic path of the projectile.

b) How high above the river will the projectile be where it crossesi. the right riverbank, x=2?ii. The left riverbank, x=0?

c) Approximately where will the projectile be when y=130?

9. In the specifications for a solar dish for heating purposes, the depth (in centimeters) is given by the following equation.

1.10d² - 10.66d – 40.32 = 0Find the insulation thickness for the piping.

Once you think you have successfully COMPLETED ALL of the word problems check with the answer key near the end. If you feel you are ready for an even bigger challenge the keep going so we can test what you have learned. If you are still having a little difficulty you may want to go over them once more!

TEST YOUR ABILITY ON QUADRATIC FUNCTIONS

This test will prove to you and others that you have mastered the task to solve word problems and Quadratic problems. We hope you have learned lots and have enjoyed your learning experience.

Good Luck!

1. State the quadratic equation. (2 points)

2. Find the roots of the following questions, using the quadratic formula. (2 points each)

A) x2 – 4x –1 = 0 B) 2x2 – 3x – 1 = 0

C) 3x2 – 5x – 1 = 0 D) 2x2 –5x + 1 = 0

3. Find the zeros using the quadratic equation. (2 points each)

A) x2 – 5x – 6 = 0 B) x2 – 7x + 12 = 0

C) 2x2 + x – 1 = 0 D) 3x2 + 8x –3 = 0

4. Find the x intercepts (4 points)

A) 9x2 – 12x + 4 = 0 B) 4x2 – 4x + 1 = 0

5. For Bay Park, a landscaper wishes to plant a boundary of tulips within a rectangular garden with dimensions 18m by 12m. To obtain a pleasing look the area of the tulip border should be half of the area of the garden. How wide should the border be?

6. The width (in meters) for the most efficient wind tunnel is given by the following equation:

Wz + 1.40w – 7.35 = 0

Solve the equation to obtain the width.

7. In the specifications for a solar dish for heating purposes, the depth (in centimeters) is given by the following equation:

1.10d2 – 10.66d – 40.32 = 0

Find the required depth.

8. By experiments it is found that the most efficient pedestrian tunnel is given by…

Y= -0.04x2 + 0.88x – 0.84

…where the floor of the tunnel is along the x-axis. Determine the height of the tunnel.

CONGRATUALTIONS! You have successfully finished and are now officially a whiz at quadratic equations! Well Done!

ANWSER KEY TO PRACTICE PROBLEMS(solving for x using the quadratic formula)

1. 3x²-5x-1=0 ___________

X=-(-5)+ or- √(-5)²-4 (3) (-1)2(3)

X= -0.18X= 1.85

2. 15x²+26x+12=0 _____________

X= -26 + or-√(26)²-4 (15) (12)2(15)

There is no solution to this problem due to the fact that our number inside the square root is a negative (-44).

3. 6x²-27x+20=0 ____________

X= -(-27) +or-√(-27)²-4 (6) (20) 2(6)

X= 3.56X=0.935

4. x²-2x+5=0 __________

X= -(-2) +or-√(-2)-4 (1) (5)2

No solution due to a negative square root.

5. 4x²-24=10x4x²-10x-24=0

_____________X= -(-10)+ or-√(-10)²-4 (4) (-24)

2(4)

X=-1.5X= 4

6. 6x²-5x=46x²-5x-4=0

___________X=-(-5)+ or-√(-5)²-4 (6) (-4)

2(6)

X=-0.5X= 1.33

7. 9-6x=-x² x²-6x+9=0

________ X= -(-6) +or-√(-6)²-4 (9)

2 X= 3

8. 2(x+4)=x²-x²+2x+8=0

__________X= -2 +or-√(2)²-4 (-1) (8)

2(-1)

X= 4X=-2

9. (2x+1)²-(2x+1)=x²-x3x²+3x+2=0

__________X= -3 +or-√(3)²-4 (3) (2)

2(3)No solution due to a negative square root.

10. (1-x) (x+4) + 2 (2x+1)²=63

5x²-1x-4=0 3 3 3

________________ X= - (-1/3) +or-√(-1/3)²-4 (5/3) (-4/3) 2(5/3)

X=-4/5 X= 1

11. 3x= 2 +4

x 3x²-4x-2=0

____________ X= - (-4) +or-√ (-4)²-4 (3) (-2)

2(3)

X=-0.39 X= 1.72

ANWSER KEY TO PRACTICE WORD PROBLEMS!

1. (60+x) (40+x) = 2815m²

x² + 100x –415 = 0 ___________

X= -100 +or-√(100)²-4 (-415)2

X= 4m The canopies should extend 4m to make them just the right length.

2. (50m + x) (27m + x) = 3835m² x² + 77x – 2485 = 0

____________ X= -77 +or- √(77)²-4 (-2485)

2

X= 24.5mThe parking strip is 24.5m wide.

3. (15m + x) (8m + x) = 240m²x² + 23x – 120 = 0 ___________X= -23 +or-√(23)²-4 (-120) 2

X=4.4m 4.4m was added onto one side and one end of the pool.

4. 50 = 1.50 (x-1) x 1.5x² - 1.5x – 50 = 0

_________________ X= -(-150) +or-√(-1.50)²-4 (1.50) (-50)

2(1.50) X= 6.3There were 6.3 friends at a movie. Al though it is not possible to have 0.3 of someone, this would explain the different prices for the tickets.

5.600 = 2.50 (x-1)x

2.50x² - 2.50x – 600 = 0 __________________

X= -(-2.50) +or-√(-2.50)²-4 (2.50) (-600) 2(2.50)

X= 16 There were 16 people at dinner.

6. We know that the area equals length multiplied by the width. Therefore the area we are trying to find will be the length of the castle add 2x (2 because the moat extends from all four walls, two along the length and two along the width) multiplied by the width of the castle add 2x.We also know that the total area will equal 16000m2. So our equation will look like this…

(150m+2x)(75m=2x)=16000m2

And when we expand this equation we get…

4x2+225x-4750=0

And using our formula to solve for x we get…

X = -225± 225 2 -4(4)(-4750) 2(4)We now solve for x ____________

X= -225± 50625-(76000) 16

______ ______x=-225- 126625 x=-225+ 126625

16 16x=-36.3m x=8.18m

Since we know a moat cannot have negative meters the answer is 8.18meters.

7. We know Pythagorean theorem: a2+b2= c2 therefore we know the original rectangles diagonal…

(8cm)2+ (6cm)2= c2

____

c=100

c=10cm

So we also know that:(8cm)2+ (6cm+x)2=(10+7)2

x2+ 12x- 189=0

Now just plug it in to solve __________

X=-12± (12) 2 -4(-189) ________

X=-12± 144-(-756) 2

___ ___X=-12- 900 X=-12+ 900

2 2

X=-21cm X=9cm

Because distance cannot be negative the answer is 9cm.

8. a) –80x² + 120x +610 = 0 b) i.-80(2²) + 120 (2) + 610 = y y=530m

ii. –80(0²) + 120 (0) + 610 = yy=610m

c)-80x² + 120x +610 = 130 -80x² + 120x +610 – 130 = 0 -80x² + 120x + 480 = 0

_______________ X= -120 +or-√(120)²-4 (-80) (480)

2(-80)

X= 3.31 X=-1.31Because the plain is descending it will mw –1.31 meters.

9. 1.10d² -10.66x – 40.32 = 0 ____________________ X= -(-10.66) +or-√(-10.66)²-4 (1.1) (-40.32)

2(1.1) X=-2.88 X=12.57The depth will be 12.57 meters.

ANSWER KEY TO: TEST YOUR ABILITY ON QUADRATIC FUNCTIONS

______Question1: X = -b + or - b 2 - 4ac

2a

Question 2: A) A = 1 B = -4 C = -1

________X = -(-4) + or - -4 2 4(1)(1)

2(1) _____

= 4 + or - 16 – 4 2

__ = 4 + or - 12

2

= 4 + 3.6 2 or x= 4 – 3.6

2

= 3.8 or 0.2

B) A = 2 B = 3 C = -1

__________X = -(-3) + or - -3 2 – 4(2)(-1)

2(2) ____

= 3 + or - 9 + 8 4 __

= 3 + or - 17 4

= 3 + 4.1 or 3 – 4.14 4

= 1.8 or -0.3

C) A = 3 B = -5 C = -1

__________X = -(-5) + or - -5 2 – 4(3)(-1)

2(3) ______

= 5 + or - 25 + 12 6 __

= 5 + or - 37 6

= 5 + 6.1 6 or 5 – 6.1

6

= 1.8 or -0.2

D) A = 2 B = -5 C = 1 __________

X = -(-5) + or - -5 2 – 4(2)(1) 2(2) _____

= 5 + or - 25 – 8 4 __

= 5 + or - 17 4

= 5 + 4.1 4 or 5 – 4.1

4

= 2.8 or 0.2

Question 3: A) A = 1 B = -5 C = -6 __________

X= -(-5) + or - -5 2 – 4(1)(-6) 2(1) ______

= 5 + or - 25 + 24 2 __

= 5 + or - 49 2

= 5 + 7 2 or 5 – 7

2

= 6 or -1

B) A = 1 B = -7 C = 12 ___________

X = -(-7) + or - -7 2 – 4(1)(12) 2(1)

______ = 7 + or - 49 – 48

2 = 7 + 1

2 or 7 – 1 2

= 4 or 3

C) A = 2 B = 1 C = -1 _________

X = -(-1) + or - 1 2 –4(2)(-1) 2(2) ___

= -1 + or - 1-8 4

Impossible can’t take the square root of a negative #.

D) A = 3 B = 8 C = -3

__________X = -(-8) + or - 8 2 – 4 (3)(-3)

2(3) ______

= -8 + or - 64 – 36

6 __

= -8 + or - 28 6

= -8 + 5.3 or 8 – 5.3 6

= -0.45 or -2.2

Question 4: A) A = 9 B = -12 C = 4 ___________

X = -(-12) + or - -12 2 – 4(9)(4) 2(9)

________ = 12 + or - 144 – 144

18

= 1218

= 0.66

B) A = 4 B = -4 C = 1 __________

X = -(-4) + or - -4 2 – 4(4)(1) 2(4) ______

= 4 + or - 16 – 16 8

= 48

= ½

Question 5: A = 1 B = -15 C = 27 ________

X = 15 + 225 – 108 2 ___

= 15 + 117

2

= 15 + 10.82 2 or 15 – 10.82

2

= 12.91 or 2.09

Question 6: A = 1 B = 1.40 C = -7.35

________________X = -1.4 + or - (1.40) 2 – 4(1)(-7.35)

2(1) _________

= -1.4 + or - 1.96 + 29.4 2 ____

= -1.4 + or - 31.36 2

= 1.4 + 5.6 2 or -1.4 – 5.6

2

= 2.1 or -3.5

Question 7: A = 1.10 B = -10.60 C = -40.32 _____________________

X = 10.66 + or - (-10.66) 2 – 4(1.10)(-40.32) 2(1.10) ______________

= 10.66 + or - 113.63 + 177.408 2.2 ___

= 10.66 + or - 291 2.2

= 10.66 + or – 17.0 2 or 10.66 – 17.0

2

= 13. 83 or -3.17

The depth of the solar dish is 13.83 cm (approximately).

Question 8: A = -0.4 B = 0.88 C = -0.84 ___________________

X= -0.88 + or - (0.88) 2 – 4(-0.04)(-0.84) 2(-0.04) _____________

= -0.88 + or - 0.7744 – 0.1344 0.08 ___

= -0.88 + or - 0.64 -0.08

= 0.88 + 0.8 -0.8 or 0.88 – 0.8

-0.8

= 1 or 21

The height of the tunnel is 20.

BIBLIOGRAPHY1.http://ppl.nhmccd.edu/~hkocurek/studyskills/wordproblems.html2.http://www.sosmath.com/Look under: algebra, quadratic formula and quadratic word problems3.http://www.mathworld.com/

Look under: algebra, quadratic formula, applications of and word problems

4.http://www.purplemath.com/modules/quadform.htm5.Pre-Calculus Mathematics One. David W. McKillop, Georgina V. Kelly. Nelson Canada, Scarborough Ontario. 19992 p. 89-90, 105-106