Module - Quadratic...

LAST REVISED November, 2008

Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved.

Algebra Module A60

Quadratic Equations - 1

Module A60 − Quadratic Equations - 1

Quadratic Equations - 1 Statement of Prerequisite Skills Complete all previous TLM modules before completing this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale

Why is it important for you to learn this material? The quadratic equation is one of the most common equations encountered in the technologies. Its applications include projectile motion, architectural problems, electrical circuits, problems in mechanics, and design. This unit will introduce the student to many different strategies for manipulating and solving quadratic equations.

Learning Outcome

When you complete this module you will be able to… Solve quadratic equations using a variety of methods.

Learning Objectives 1. Rewrite algebraic expressions in the standard quadratic form and state the values of

a, b, and c. 2. Solve pure quadratic equations. 3. Solve quadratic equations by factoring. 4. Solve a quadratic equation by completing the square. 5. Solve a quadratic equation by formula. 6. Solve a quadratic using any method.

Connection Activity Consider the following triangle:

x

x + 1

6

Your previous experience with the Pythagorean theorem tells you that x2 + (x + 1) 2 = 62. Simplifying this equation we get x2 + x2 + 2x + 1 = 36 or 2x2 + 2x – 35 = 0. This is a quadratic equation and solving the equation will allow you to solve for the length of the sides of the triangle (x).

1

http://www.microsoft.com/windows/ie/default.asp

http://www.macromedia.com/shockwave/download/download.cgi?P1_Prod_Version=ShockwaveFlash

Module A60 − Quadratic Equations - 1 2

OBJECTIVE ONE When you complete this objective you will be able to… Rewrite algebraic expressions in the standard quadratic form and state the values of a, b, and c.

Exploration Activity Given that a, b and c are constants and a ≠ 0, the equation:

ax2 + bx + c = 0 is called the Standard Form of a quadratic equation.

Note: When identifying a, b and c, include the signs if they are negative.

EXAMPLES 1. 2x2 + 3x + 7 = 0 ← This is in the form of ax2 + bx + c = 0 and a ≠ 0. Therefore it is a

quadratic equation with a = 2, b = 3, c = 7. 2. 4x2 − 7x + 9 = 0 ← Quadratic: yes or no? Yes, with a = 4, b = −7, c = 9. 3. x3 − 9x + 8 = 0 ← Quadratic: yes or no? No, this is not a quadratic because of the

presence of the x3 term, and also because a = 0. 4. 2x2 = x2 − 16x + 64 ← This is not in the general quadratic form, but it can be

rearranged. Move all the terms on the right hand side of the equation to the left hand side, by subtracting or adding. Then get:

2x2 − x2 + 16x − 64 = 0

which simplifies to: x2 + 16x − 64 = 0. This is a quadratic with a = 1, b = 16, and c = −64.

5. x(x2 − 7) = x3 + x2 + 4x ← This equation must be expanded and simplified first.

Drop brackets: x3 − 7x = x3 + x2 + 4x. Now collect like terms to get: x2 + 11x = 0

Here a = 1, b = 11, and c = 0.


Experiential Activity One Determine which of the following equations are quadratics. If the standard form ax2 + bx + c = 0 can be obtained, identify a, b and c. 1. x2 − 7x = 4 2. 3x2 = 5 − 9x 3. x2 = (x − 1)2 4. 2x2 − x = 2x(x + 8) 5. (x + 2)2 = 0 6. x(x2 − 1) = x3 + x2 7. x2 = x(1 − 6x) 8. x(1 + 2x) = 3x − 2x2 9. x2(1 − x) = 4 10. x(x2 + 2x − 1) = 0 11. 3x(x2 − 2x + 1) = 1 + 3x3 12. x(x3 + 6) − 1 = x2(x2 + 1)

13. 640+=

xx 14. 030101 2 =++

xx

15. ( ) ( )3

34235 −

=−−−− xx

xx

Show Me. 16. 7

14

22

++−

=+−

xx

xx

Experiential Activity One Answers 1. a = 1; b = − 7; c = −4 2. a = 3; b = 9; c = −5 3. Not a quadratic (a = 0) 4. Not a quadratic (a = 0) 5. a = l; b = 4; c = 4 6. a = 1; b = 1; c = 0 7. a = 7; b = −1; c = 0 8. a = 4; b = −2; c = 0 9. Not a quadratic 10. Not a quadratic 11. a = 6; b = −3; c = 1 12. a = 1; b = −6; c = 1 13. a = l; b = −6; c = −40 14. a = l; b = l0; c = 30 15. a = 4; b = −38; c = 75 16. a = 7; b = 20; c = 8

3


OBJECTIVE TWO When you complete this objective you will be able to… Solve pure quadratic equations.

Exploration Activity A quadratic equation in which the x term is missing, i.e. in which b = 0, is called a pure quadratic equation.

EXAMPLES 1. R2 − 49 = 0 ← Note: a = 1, b = 0, c = −49 2. 3x2 = 12 ← Note: a = 3, b = 0, c = −12 Solving a Pure Quadratic Equation.

EXAMPLE 1 Solve: x2 − 16 = 0. SOLUTION: Isolate x by adding 16 to both sides.

x2 − 16 + 16 = 0 + 16 x2 = 16

Now take the square roots of both sides.

162 =x

and x = ± 4

Check: x2 − 16 = 0 i) if x = 4 16 − 16 = 0 ii) if x = −4 (−4)2 − 16 = 0 16 − 16 = 0

4


EXAMPLE 2 Solve 5A2 − 89 = 91. SOLUTION: Isolate A. Add 89 to both sides to get:

5A2 − 89 + 89 = 91 + 89 5A2 = 180

Now divide by 5,

51802 =A

A2 = 36 Take the square root of both sides to get:

A = ±6 Check:

i) if A = 6 5A2 − 89 = 91 5 × 62 − 89 = 91 91 = 91

ii) if A = − 6

5A2 − 89 = 91 5 × (−6)2 − 89 = 91 91 = 91

5


Experiential Activity Two Solve the following and check your answers. 1. x2 − 25 = 0 2. x2 − 49 = 0 3. x2 + 36 = 225 4. x2 − 0.25 = 0 5. 5x2 − 180 = 0 6. x2 − 0.0004 = 0.0012

7. 0121

92 =−x 8. 49x2 − 144 = 0

9. 5135 2 =x 10. 5x2 − 0.0308 = 0.0817

11. 2(m +1 ) − m(m − 3) − 5m = 0 12.

331

33

9282 +

−+−

−+

=− x

xxx

x

Show Me.

Experiential Activity Two Answers Note: Answers are not supplied for these exercises in the hope that the student will get in the habit of CHECKING his/her answers to see if they are correct. Decimal answers may not always check exactly because of the number of decimal places taken. “Close” in these cases is good enough. Try to avoid using decimals and use fractions instead.

6


7

OBJECTIVE THREE When you complete this objective you will be able to… Solve quadratic equations by factoring.

Exploration Activity If the polynomial ax2 + bx + c can be factored easily, then there is a simple method for solving the equation ax2 + bx + c = 0. The method is based on the Zero−Product theorem.

Zero−Product Theorem: If the product of two real numbers, a and b, is zero, then one of the factors of the product, a or b must be zero. That is, a · b = 0 if a = 0 or b = 0.

This theorem can be extended to algebraic expressions. For instance if (x + 3)(x − 2) = 0 this is possible if and only if either (x + 3) = 0, or (x − 2) = 0 Solving quadratics by factoring.

EXAMPLE 1 Solve x2 + 5x + 4 = 0 for x and check your answer. SOLUTION: To factor we must find 2 integers that have a product of +4, and a sum of +5. 4 and 1 seem likely numbers to try so we have:

(x + 4)(x + 1) = 0 Now to solve, we set each factor to zero and solve the equation. so x = −4 and x = −1 Check:

i) if x = −4 x2 + 5x + 4 = (−4)2 + 5(−4) + 4 = 16 − 20 + 4 = 0. Thus (−4) checks.

ii) and for x = −1

x2 + 5x + 4 (−1)2 + 5(−1) + 4 = 0. Thus (−1) checks.


EXAMPLE 2 Solve 6x2 = 5 − 7x and check. SOLUTION: Arrange the equation into proper format.

6x2 + 7x − 5 = 0

Factor: use a trial and error method. (2x − 1)(3x + 5) = 0

Set each factor equal to zero and solve.

2x − 1 = 0 and 3x + 5 = 0

x =21

and x = 35

−

Check:

i) if x = 21

6x2 = 5 − 7x 6( 2

1 )2 = 5 − 7( 21 )

23

= 23

ii) if x= 3

5− 6x2 = 5 − 7x 6( 3

5− )2 = 5 − 7( 35− )

350

= 3

50

16.67 = 16.67 HINT: When you are trying to find the factors of a quadratic such as the previous example where the coefficient of the x2 term has several combinations of factors, i.e. 6 = 6 × 1 and 6 = 3 × 2, try the combination in which the factors are close in size, i.e. try the 3 and 2. Taking the extremes, i.e. 6 and 1 may work but more often the "closer" factors lead you into the solution of the problem quicker. This idea also works when looking at the constant term. For instance if c = 24, do not choose 24 and 1 as factors, but start with 6 and 4 and work from there. (Good luck, as there is no easy way out. Practice, practice, practice. ) Also, look for the patterns in numbers, always keeping in mind the middle term.

8


9

Experiential Activity Three Solve the following quadratics by factoring and check your answers. 1. x2 + x − 2 = 0 2. x2 − 5x + 6 = 0 3. t2 + t = 12 4. x2 = 6x + 7 5. 3x2 + 5x − 2 = 0 6. 4x2 − 7x + 3 = 0 7. 2x2 + 3x = 2 8. 3y2 = 8 − 10y 9. 6x2 + 13x − 5 = 0 10. 6x2 − 7x − 20 = 0 11. 6n2 − n = 2 12. 2v2 + 30 = 17v 13. 5x2 + 4 = 21x 14. 9p2 + 20 = −27p Show Me. 15. 2x2 = 9 + 7x 16. 6x2 + 7x = 10 17. x2 − 4x + 4 = 0 18. x2 + 9 = 6x 19. x2 − 5 = 0 20. 2x2 + 7x = 0 21. 8m2 + 24m = 14 22. 9x2 + 30x + 25 = 0 23. 4x2 + 49 = 28x 24. 4x2 + 4x + 1 = 0 25. t2 = 3t 26. 3r2 = 21r 27. 8m + 3 = 3m2 28. 3 − x = 4x2 29. x2 − 4a2 = 0 (a is constant) 30. 2x2 − 8ax + 8a2 = 0 (a is constant)

Experiential Activity Three Answers Answers are not supplied for these exercises in the hope that the student will get in the habit of CHECKING his/her answers to see if they are correct. Decimal answers may not always check exactly because of the number of decimal places taken. “Close” in these cases is good enough. Try to avoid using decimals and use fractions instead.


OBJECTIVE FOUR When you complete this objective you will be able to… Solve a quadratic equation by completing the square.

Exploration Activity Some quadratics are not readily solved by factoring, but can be solved by a method known as completing the square. To solve by completing the square: Step 1: make sure the coefficient of the x2 term is +1. If it isn't +1 then divide each term

of the equation by a number that will make the coefficient of the x2 term +1. Step 2: move the c term to the right hand side. Step 3: add a constant to both sides which makes the left hand side a perfect square. This

constant is the square of 1/2 of the coefficient of the x term. Step 4: factor the left hand side, as it now a perfect square. Step 5: take the square root of both sides. Step 6: solve for x. NOTE: Follow Example 1 and keep these steps in mind.

EXAMPLE 1 Solve x2 – l0x − 20 = 0 by completing the square. SOLUTION: To complete the square: Check to see if the coefficient of x2 = 1. It is, proceed, Move the constant −20 to the other side of the equation by addition

x2 − 10x = 20

Now take 21 the coefficient of the x term which is −10, and square it, getting 25. Add this

25 to both sides:

x2 − 10x + 25 = 20 + 25

x2 − 10x + 25 = 45 10


Factor the left side as a perfect square:

(x − 5)(x − 5) = 45 (x − 5)2 = 45

Now the take square root of both sides:

455 ±=−x x − 5 = ± 53 x = +5 ± 53 x = 11.71 and −1.71

Exact form

Approximate form Check:

i) if x = 11.71 (11.71)2 = 10(11.71) − 20 =137.1 − 117.1 − 20 = 0

ii) if x = −1.71

(−1.71)2 − 10(–1.71) − 20 = 0

EXAMPLE 2 Solve 3x2 − x −1 = 0 by completing the square. SOLUTION: To complete the square: Check to see if the coefficient of x2 = 1. It isn't, so divide through by 3.

We get: 031

312 =−− xx

31

312 =− xx

Now the constant to be added to both sides is obtained by taking

61

31

21

31 of

21

−=−⋅=− .

squaring it:361

61 2

=⎟⎠⎞

⎜⎝⎛−

11


So we get:

3613

61

3613

61

61

361

31

361

31

2

2

=⎟⎠⎞

⎜⎝⎛ −

=⎟⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛ −

+=+−

x

xx

xx

Now take the square root of both sides

1 136 36

1 136 6

1 136

0.7676 and 0.4343

x

x

x

x x

− = ±

= ±

±=

= = −

Check:

i) if x = 0.7676 3x2 − x − 1 = 0

ii) if x = −0.4343 3x2 − x − 1 = 0

12


Experiential Activity Four Solve by completing the square and check your answers. 1. x2 − 8x + 12 = 0 2. a2 − 4a − 45 = 0 3. E2 − 15E + 54 = 0 4. x2 + 5x + 6 = 0 5. i2 − 27i = −50 6. 63 − a2 − 2a = 0 7. θ2 + 2 = 3θ 8. e2 − 6 = e 9. M2 = 22M + 48 10. 24E2 = 2E + 1 11. 3 + θ = θ2 − 3 12. 17I − 42 = I2 + 2I − 16

13. 460+

φ=φ 14. 035121 2 =++

ff Show Me.

15. ( ) ( )2

42347 −

=−−−− RR

RR 16. 6

22

11

−+−

=+−

ZZ

ZZ

Experiential Activity Four Answers Answers are not supplied for these exercises in the hope that the student will get in the habit of CHECKING his/her answers to see if they are correct. Decimal answers may not always check exactly because of the number of decimal places taken. “Close” in these cases is good enough.

13


OBJECTIVE FIVE When you complete this objective you will be able to… Solve a quadratic equation by formula.

Exploration Activity The formula for solving a quadratic equation is found by completing the square for the general quadratic equation. Study the following and be prepared to develop this formula on a supervised test. ax2 + bx + c = 0 To Complete the Square

02 =++acx

abx ← Coefficient of x2 must be one, therefore divide each term by a.

acx

abx −=+2 ← Move constant to the right.

2

2

2

22

44 ab

ac

abx

abx +−=++ ←

To find the constant to be added to both sides:

ab

ab

2 of

21 take = and square it to get 2

22

42 ab

ab

=⎟⎠⎞

⎜⎝⎛

ac

ab

abx

abx −=⎟

⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ + 2

2

422 ← Factor left hand side.

ac

ab

abx −=⎟⎠⎞

⎜⎝⎛ + 2

22

42

ac

ab

abx −±=+ 2

2

42 ← Take the square root of both sides; then solve for x.

ac

ab

abx −±

−= 2

2

42

2

2

44

2 aacb

abx −±

−= ← Simplify algebraically.

aacbbx

aacb

abx

a

acbabx

24

24

2

4

42

2

2

2

2

−±−=

−±

−=

−±

−=

14


The formula:

aacbbx

242 −±−

=

is known as the quadratic formula. It is extremely useful!

Instead of attempting to solve a quadratic equation by factoring or completing the square we now can use the quadratic formula.

EXAMPLE 1 Solve 5x2 + 2x − 3 = 0 using the quadratic formula. SOLUTION: Identify a, b, and c: a = 5, b = 2, c = −3 Now substitute into the formula:

( )( )( )( )( )

1082 and

1082

1082

10642

106042

5235442

242

−−=

+−=

±−=

±−=

+±−=

−−±−=

−±−=

x

aacbbx

so: x = 0.6 and x = −1

Check:

i) if x = 0.6 5x2 + 2x − 3 = 0

ii) if x = −1 5x2 + 2x − 3 = 0

15


EXAMPLE 2

Using the quadratic formula, solve for R in RR

25

3=

−

SOLUTION: Arrange the equation in standard quadratic form. Multiply both sides by (5 − R),

( ) ( )

( )RR

RRRR

−=

−=−−

523

5255

3

Expand and collect like terms,

3 = l0R − 2R2

2R2 − l0R + 3 = 0 Identify a , b, and c.

a = 2, b = −l0, and c = 3.

aacbbR

24 Use

2 −±−=

( ) ( ) ( )( )( )

472.810 and

472.810

472.810

47610

223241010 2

−=

+=

±=

±=

−−±−−=

RR

R = 4.68; and R = 0.320

16


Check:

i) if R = 4.68

RR

25

3=

−

9.36 = 9.36

ii) if R = 0.320

RR

25

3=

−

0.64 = 0.64

17


Experiential Activity Five Solve the following equations by using the quadratic formula and check your answers. 1. θ2 = 4 − 3θ 2. x2 + 7 = 18 3. 2I + 35 = I2 4. a2 − 4a + 3 = 0 5. 15 − 14q = 8q2 6. 3I2 − 7I + 2 = 0 7. 5 = 6Z2 − 3Z 8. 5(R + 2) = 2R(R − 1)

9. 01224 2 =−−mm

10. 141322111−=+

III

11. 03

1214

11

1 =−

−−−

RRR 12. 1

23

32

−−

=+ xx

13. 4

221

37

−−

=−− xx

14. ( )

132

336

2 =++

−+ I

II

Show Me.

15. 43162157 2

3

−−

=+iii 16.

EE

EE

725

214

2 +−=−−

Experiential Activity Five Answers Answers are not supplied for these exercises in the hope that the student will get in the habit of CHECKING his/her answers to see if they are correct. Decimal answers may not always check exactly because of the number of decimal places taken. “Close” in these cases is good enough.

18


19

OBJECTIVE SIX When you complete this objective you will be able to… Solve a quadratic using any method.

Exploration Activity Solve a quadratic by any method. It is important that the student be able to use any of the 3 methods discussed for solving a quadratic equation.

Experiential Activity Six For each of the following quadratic equations solve by: 1. Factoring (if possible) 2. Completing the square 3. Quadratic formula Check your answers. 1. x2 − x − 6 = 0 2. x2 + 2x − 15 = 0 3. 2x2 + x − 1 = 0 4. 3y2 − 5y + 2 = 0 5. x2 + 5x + 3 = 0 6. x2 − 3x − 1 = 0 7. s2 = 4s + 2 8. n2 − 2n = 6 9. 4t2 = 8t − 3 10. 3x2 = x + 10

Experiential Activity Six Answers Answers are not supplied for these exercises in the hope that the student will get in the habit of CHECKING his/her answers to see if they are correct. Decimal answers may not always check exactly because of the number of decimal places taken. “Close” in these cases is good enough.

Practical Application Activity Complete the Quadratic Equations - 1 assignment in TLM.

Summary This module introduced a variety of methods for solving quadratic equations.

Module - Quadratic...

Documents

Transcript of Module - Quadratic...