Survey of Mathematics - University of Hawaiilittle/m100chI.pdfSurvey of Mathematics Adolf Mader, PSB...

Survey of Mathematics

Adolf Mader, PSB 308B, Tel. 956-6813

[email protected]

Monique Chyba, K311, Tel. 956–8464,

[email protected]

CHAPTER 1

Solving equations

1. Linear equations and linear systems

How did mathematics come about and what does it deal with? We consider someexamples of early, but already highly developed mathematics.

Exercise 1.1. The following problem appears on a Babylonian clay tablet of around300 BC. Solve.There are two fields whose total area is 1800 square yards. One produces grain at therate of 2/3 of a bushel per square yard while the other produces grain at the rate of1/2 a bushel per square yard. If the total yield is 1100 bushels, what is the size ofeach field?

Exercise 1.2. In the Chinese text “Nine Chapters on the Mathematical Arts” writ-ten around 150 BC, there appears the following problem.There are three types of corn, of which three bundles of the first, two bundles of thesecond, and one of the third make 39 measures. Two of the first, three of the secondand one of the third make 34 measures. And one of the first, two of the second, andthree of the third make 26 measures. How many measures of corn are contained inone bundle of each type? Solve.

Exercise 1.3. (Rhind Papyrus 1650 BC) Think of a number, and add 23

of this

number to itself. From this sum subtract 13

of its value. Say the answer is 10. Then

take away 110

of this 10, giving 9. Then this was the number first thought of.

Exercise 1.4. (Rhind Papyrus 1650 BC) A quantity and 23

of it and 12

of it and 17

of it added together gives 33. What is the quantity?

Exercise 1.5. (Rhind Papyrus 1650 BC) Divide 100 loaves among five men so thatthe sum of the three largest shares is seven times the sum of the two smallest shares.Comment. This problem is ambiguous as stated. Is it possible to solve it withoutcutting up any loaf? If not, how much cutting is unavoidable?Solve the problem assuming that

(1) the two smallest shares are equal and the three largest shares are equal;(2) the difference between some share and the next bigger one is a fixed quantity.

(This is the actual assumption of the papyrus.)

3

4 1. SOLVING EQUATIONS

Definition 1.6. A linear equation with unknown x is an equation that can besimplified by algebraic manipulations to the form

(1.7) Ax = B, where A, B are numbers.

Proposition 1.8. Let a linear equation (1.7) be given. Then there are three cases.

(1) A = B = 0. In this case every number x is a solution.(2) A = 0 and B 6= 0. In this case there is no solution.(3) A 6= 0. In this case there is the unique solution x = B/A.

Example 1.9. Solve

2 + 5x− 3x + 13x = 5− 2

3x + 3x− 3.

Collecting terms we get2 + 7

3x = 2 + 7

3x.

Moving terms we get (73− 7

3

)x = 2− 2

or0x = 0

Every number x solves this equation.

Example 1.10. Solve

2 + 5x− 3x + 13x = 5− 1

3x + 3x.


3x = 5 + 8

3x.


3

)x = 5− 2

or−1

3x = 3

The unique solution is x = −9.Check: Left hand side (LHS): 2+5(−9)−3(−9)+ 1

3(−9) = 2−45+27−3 = 29−48 =

−19.Right hand side (RHS): 5 − 1

3(−9) + 3(−9) = 5 + 3 − 27 = −19. So x = −9 solves

the equation.

Example 1.11. Solve

2 + 5x− 3x + 13x = 5− 2

3x + 3x.


3x = 5 + 7

3x.


3

)x = 5− 2

1. LINEAR EQUATIONS AND LINEAR SYSTEMS 5

or0x = 3

There is no solution.

Exercise 1.12. Solve and check your answer.

(12

+ 3)x− 13x = x + 1.

Exercise 1.13. Solve and check your answer.12x− 1

3x + 1

2= 3x + 1.


(12

+ 3)x− 13x = x + 13.


x + 2x + 3x + 4x + 5x = x− 2x + 3x− 4x + 5x.

Exercise 1.16. Solve and check your answer.12x + 1

3x + 1

4= 1

12x + 1.


x + 12x + 1

4x + 1

8x = 3.


3x + 1

4x = x + 1

12x + 1.


3x + 1

4x− 1

12x = x.

1.20. Equality. For all a, b, c,

(1) a = a(2) If a = b, then b = a.(3) If a = b = c, then a = c

(a = b = c is shorthand for a = b AND b = c)(4) Substitution Rule. If a = a′, then in any expression, a can be replaced by

a′.

1.21. The Arithmetic of Numbers. Numbers can be added and multiplied. Thefollowing are the practical rules for the arithmetic of numbers.

(1) In every finite sum, the summands can be rearranged in any desired orderand the additions can be executed in any order.

(2) In every finite product, the factors can be rearranged in any desired orderand the multiplications can be executed in any order.

(3) There are unique special numbers 0, 1, with the property that, for all x,0 + x = x + 0 = x, and 1 · x = x · 1 = x.


(4) For every number x there is a unique number −x, the negative or additiveinverse of x, such that x + (−x) = 0.

(5) For every number x 6= 0 there is a unique number 1/x = x−1, the multi-plicative inverse or reciprocal of x, such that x · 1

x= 1.

(6) To compute the product of a sum with another sum one has to add up allproducts of elements in the one sum with elements in the other sum.

(7) Subtraction. x− y := x + (−y).(8) Division. x÷ y := xy−1 = x/y = x · 1

ywhere y must be nonzero.

Example 1.22.

(1) 2 + 4 + (−5) + 6 + (−2) = (2 + (−2)) + (4 + 6) + (−5) = 10− 5 = 5.(2) 1

3(−5)2

361

5= (6 · 1

3)((−5)1

5)2

3= 2 · (−1) · 2

3= −4

3.

(3) Let x = 13

+ 25

be given. Then x−1 =(

5+615

)−1= (11

15)−1 = 15

11.

(4) 23

+ −23

= 0.

(5) −23· 3−2

= 1.(6) (1 + x)(a + b + c) = a + b + c + xa + xb + xc.

Exercise 1.23. Verify the following (important) identities by multiplying out the lefthand sides..

(1) (a + b)2 = a2 + 2ab + b2.(2) (a + b)3 = a3 + 3a2b + 3ab2 + b3.(3) (a− b)(a + b) = a2 − b2.(4) (a− b)(a2 + ab + b2) = a3 − b3.

The following formulas are consequences of 1.21 and may help.

Lemma 1.24.

(1) x · 0 = 0 · x = 0.(2) −x = (−1) · x.(3) −0 = 0.(4) (−1)(−1) = 1.(5) −(−x) = x.(6) If x + y = x + z then y = z.(7) xy = 0 if and only if x = 0 or y = 0.(8) (x−1)−1 = x.

Exercise 1.25. Answer the following questions. Use the (Egyptian) approximation227

to π, i.e., use π = 227.

(1) The volume V of a sphere with radius r is given by the formula V = 43πr3.

(a) If r = 4, what is V ?(b) If V = 3

4what is r?

(2) The volume V of a pyramid of height h with rectangular base with sides oflength a and b is given by the formula V = 1

3abh.


(a) If a = 10, h = 5 and V = 100, what is b?(b) If a = 10, b = 5 and V = 100, what is h?(c) If a = 10, b = 5 and h = 3, what is V ?

Exercise 1.26. Answer the following questions.

The formula that relates the measurement in centigrades C and in Fahrenheits F ofa temperature is

F = 32 +9

5C.

(1) What is the temperature in Fahrenheits that is the same as 35◦C?(2) What is the temperature in centigrades that is 104◦F?(3) For which temperature is its measurement in centigrades equal to the negative

of its measurement in Fahrenheits?(4) For which temperature is the measurement in Fahrenheits three times its

measurement in centigrades?(5) For which temperature is the measurement in Fahrenheits two times its mea-

surement in centigrades?(6) For which temperature is the measurement in Fahrenheits 1

2times its mea-

surement in centigrades?

1.27. The number systems.

(1) N = {1, 2, . . .}, the natural numbers.(2) Z = {0,±1,±2, . . .}, the integers.(3) Q = {a

b| a, b ∈ Z, b 6= 0}, the rational numbers.

(4) R = {x | x is a decimal fraction}, the real numbers.

1.28. Manipulating equations. For all a, b, c,

(1) a = b if and only if a + c = b + c.(2) If a = b, then ac = bc.(3) If ac = bc AND c 6= 0, then a = b.(4) a = b AND c 6= 0 if and only if a/c = b/c.(5) If a = b and b = c, then a± c = b± d, a · c = b · d, and, provided b = c 6= 0,

also a/c = b/d.

1.29. Solving Systems of Equations. If more than one equation is given, proceedas follows.

(1) Choose one equation to eliminate one unknown, say x, from the other equa-tions. At the end this equation is used to compute x. Ignoring theequation that has been used, the result is a system with one less equation andone less unknowns.

(2) In the next step we eliminate a second unknown, say y using a second equa-tion. This equation will later be used to compute y.


(3) Continuing in this fashion we finally have one equation that can be solved.(4) A linear system may have no solution, exactly one solution or many solutions.

Example 1.30. Solve

2x + 3y = 1

x− y = −2

Solution. We use the second equation to eliminate the unknown y. This is bestachieved by adding 3 times the second equation to the first equation. Multiplying thesecond equation by 3 we get

2x + 3y = 1

3x− 3y = −6

Adding the two equations we get

5x = 1− 6 = −5

and conclude that x = −1. We now use the second equation to compute y: y =x + 2 = −1 + 2 = 1.

Example 1.31. Solve

2x + 3y = 1

3x− 2y = −2

Solution. We use the first equation to eliminate the unknown x. This is best achievedby subtracting 2/3 times the second equation from the first equation. Multiplyingthe second equation by 2/3 we get

2x + 3y = 1

2x− 4

3y = −4

3Subtracting we get (

3 + 43

)y = 1 + 4

3

0r, equivalently,133y = 7

2

and conclude that y = 713

. We now use the first equation to compute x: x =12(1− 3y) = 1

2(1− 21

13) = 1

2· (− 8

13) = − 4

13.

Check.

2x + 3y = 2 · (− 4

13) + 3 · 7

13=−8

13+

21

13=

13

13= 1

and

3x− 2y = 3 · (− 4

13)− 2 · ( 7

13) =

−12

13+−14

13=−26

13= −2.


Exercise 1.32. The following formula expresses the connection between s, the dis-tance traveled, the starting position s0 (at time t = 0), the velocity v, and the timeelapsed t for a uniform (= constant velocity) linear motion (= along a straight line).

s = vt + s0.

0 s0s

Use the information given in the table below and compute the value of the box that ismarked by “?”. Assume that s and s0 are in miles, v is in miles/hour, t is in hours.

s v t s0

? 60 1.25 0350 80 ? 0300 ? 5 0400 ? 3 50400 50 2 ?200 55 ? 30? 45 5 25

100 ? 3 −20

Exercise 1.33. Solve and then check whether your solutions satisfy the equations.

(1)

x + 3y = 1

3x− 2y = −2

(2)

x + y = 4

3x− 2y = −2

(3)

y = 4

3x− 2y = −2

(4)

4x + 3y =1

23x− 2y = 2

(5)

0x + 0y = 1

3x− 2y = 2


(6)

x + 3y = 11

3x + y =

1

3.

Exercise 1.34. (From the Nine Chapters) Three sheaves of good crop, 2 sheaves ofmediocre crop, and 1 sheaf of bad crop are sold for 39 dou. Two sheaves of good, 3 ofmediocre, and 1 of bad are sold for 34 dou. One sheaf of good, 2 of mediocre, and 3of bad are sold for 26 dou. What is the price for a sheaf of good crop, mediocre crop,and bad crop?

Exercise 1.35. (From Bhaskara 1200 AD) Somebody owns 300 rupees and 6 horses.Another has 10 horses and a debt of 100 rupees. Both own equally much. What isthe value of a horse?

Exercise 1.36. A father and a son together are 40 years old. The father is 30 yearsolder than the son. How old are they?

Exercise 1.37. Mary is 20 years old. She is twice as old as Ann was when Marywas as old as Ann is now. How old is Ann now?

Exercise 1.38. How much copper must be added to 5 grams of 80 % gold to make60 % gold?

Exercise 1.39. At a party there are by a half more men than ladies. However, after6 men left and 6 ladies joined there were by a half more ladies than men. How manymen and women were there initially?

1.40. The Arithmetic of Fractions. In the following it is always assumed that theletters stand for integers and that the denominators are nonzero.

(1) acbc

= ab.

(2) ab· c

d= ac

bd.

(3) ad

+ bd

= a+bd

.

(4) ab

+ cd

= ad+bcbd

but this can lead to large numbers and be inefficient.(5) n is a common denominator of the fractions a

band c

dif n = bb′ and

n = dd′ for some integers b′ and d′. If so,

ab

+ cd

= ab′

n+ cd′

n= ab′+cd′

n.

(6) n = n1

if n is an integer. This shows that every integer is a rational number.(7) −a

b= (−1)a

b= −a

b= a

−b.

(8)(

ab

)−1= b

aprovided that a 6= 0.

(9)abcd

= ab

/cd

= ab÷ c

d= a

b·(

cd

)−1= a

bdc

= adbc

.

2. QUADRATIC EQUATIONS 11

2. Quadratic equations

Definition 2.1. A quadratic equation with unknown x is an equation that can besimplified by algebraic manipulations to the form

(2.2) Ax2 + Bx + C = 0, where A 6= 0, B, C are given numbers.

In order to continue we need to discuss some properties of the real number system R.

Lemma 2.3. Let x be a real number. Then x2 ≥ 0. The equation x2 = 0 has theunique solution x = 0. For a > 0, the equation x2 = a has exactly two solutions x1

and x2, one positive and the other negative, and x2 = −x1.

Definition 2.4. The square root of a number a ≥ 0, denoted√

a, is the uniquesolution ≥ 0 of the equation x2 = a.

Lemma 2.5. The (special) quadratic equation

Ax2 + C = 0, where A 6= 0

either has

(1) no solution if −C/A is negative,(2) the unique solution x = 0 if C = 0,

(3) the two distinct solutions +√−C/A and −

√−C/A if −C/A is positive.

Remark 2.6. The expression −C/A looks like a negative number, but this need notbe the case. Suppose that A = −1 and C = 2. Then −C/A = 2 is positive.

Lemma 2.7. The following formula constitutes the process of completing squares.

x2 + bx =(x + b

2

)2 − b2

4.

Proof. The formula can be verified by simplifying the right hand side of theidentity. In order to motivate the formula, we begin with the observation that

(x + u)2 = x2 + 2ux + u2

or

x2 + 2ux = (x + u)2 − u2.

The expression x2 + bx matches the left hand side of the above equation if we takeu = b/2. Substituting this expression we find that

x2 + bx =(x + b

2

)2 − b2

4.

�

Example 2.8.

(1)

x2 − 4x = (x− 2)2 − 4.


(2)

x2 + 3x = (x + 32)2 − 9

4.

(3)

5x2 − 4x = 5(x2 − 4

5x)

= 5((x− 2

5)2 − 4

25

)= 5(x− 2

5)2 − 4

5.

(4)

5x2−4x−3 = 5(x2 − 4

5x)−3 = 5

((x− 2

5)2 − 4

25

)−3 = 5(x− 2

5)2− 4

5−3 = 5(x− 2

5)2− 19

5.

Exercise 2.9. Complete the squares.

(1) x2 + 5x.(2) 2x2 + 5x.(3) 2x2 + 5x + 3.


(1) x2 − 5x.(2) 3x2 − 5x.(3) 3x2 − 5x + 4.


(1) x2 − 23x.

(2) 3x2 − 2x.(3) 3x2 − 2x + 4.


(1) x2 − 23x.

(2) 12x2 − 3

4x.

(3) 12x2 − 3

4x + 3

2.

We will now find the solution formula for an arbitrary quadratic equation. By com-pleting a square we can reduce the general case to the special case Lemma 2.5.

Proposition 2.13. The quadratic equation

Ax2 + Bx + C = 0, A 6= 0,

has

(1) the unique solution −B/2A if B2 − 4AC = 0,(2) no solution if B2 − 4AC < 0,(3) two different solutions

12A

(−B ±

√B2 − 4AC

)if B2 − 4AC > 0.

2. QUADRATIC EQUATIONS 13

Proof. Completing squares we obtain

Ax2 + Bx + C = A(x + B

2A

)2 − B2

4A+ C = A

(x + B

2A

)2 − 14A

(B2 − 4AC) .

Hence Ax2 + Bx + C = 0 is equivalent with

A(x + B

2A

)2 − 14A

(B2 − 4AC) = 0.

The general quadratic formula now follows as in Lemma 2.5.

A(x + B

2A

)2 − 14A

(B2 − 4AC) = 0.

if and only if (x + B

2A

)2= 1

4A2 (B2 − 4AC)

if and only ifx + B

2A= ± 1

2A

√B2 − 4AC

if and only ifx = 1

2A

(−B ±

√B2 − 4AC

). �

Example 2.14. Solve the quadratic equation

4x2 − 3x− 6 = 0.

Comparing with the standard form (2.2) we see that in our case A = 4, B = −3 andC = −6. Hence B2 − 4AC = 9 + 96 = 105 is positive and there are the two solutions

x1 = 18(3 +

√105), x2 = 1

8(3−

√105).


4x2 − 3x + 6 = 0.

Comparing with the standard form (2.2) we see that in our case A = 4, B = −3 andC = 6. Hence B2 − 4AC = 9− 96 = −90 is negative and there are no solutions.


4x2 − 3x− 6 + 2x = x2 + 4x− 3.

This equation must be brought into standard form first. Moving all terms to the leftwe get

3x2 − 5x− 3 = 0

Comparing with the standard form (2.2) we see that in this case A = 3, B = −5 andC = −3. Hence B2 − 4AC = 25 + 36 = 61 is positive and there are the two solutions

x1 = 16(5 +

√61), x2 = 1

6(5−

√61).

Exercise 2.17. Solve the quadratic equations.

(1) x2 + x− 1 = 0.(2) −2x2 + 3x = 0.(3) −2x2 − x + 7 = 0.


(4) −x2 − x = 10.(5) 2x2 − 4x + 5 = −x2 + 10.(6) −x2 − 5x = −5x + 3.(7) −x2 − 5x = −5x− 3.(8) 2− 4x− x2 = 0.(9) 3x− x2 + 5− x = 0.

3. Quadratic functions

The process of completing squares is also useful in the study of quadratic functions.

Definition 3.1. A quadratic function is a function f whose values or outputsf(x) for the input x are computed by means of a formula

f(x) = Ax2 + Bx + C

where A, B, C, are fixed numbers and A 6= 0.

Example 3.2. Let f be such that

f(x) = x2 − 2x + 3.

Then f is a quadratic function.For the input x = 5, the output is f(5) = 52− 2 · 5 + 3 = 25− 10 + 3 = 18. Similarly,f(0) = 3, f(1) = 2, f(−2) = 11.

Example 3.3. A farmer has 2400 ft of fence and wishes to enclose a rectangular plotalong a river, such that the river forms one side of the plot that needs no fence. Howmust the dimensions of the plot be chosen so that the plot has an area that is as largeas possible?

Solution. Let x and y be the lengths in feet of the sides of the enclosed plot suchthat x is the length of the sides perpendicular to the river and y is the length of theside parallel to the river. Then since the farmer uses all of the fencing material thathe has, we have

2x + y = 2400, or y = 2400− 2x.

Then the area A of the plot is

A = xy = x(2400− 2x) = −2x2 + 2400x

= −2(x− 600)2 − (−2)6002 = 720, 000− 2(x− 600)2.

We see that the largest possible area is 720, 000 ft2 which is obtained for x = 600ftand y = 1200 ft. �

Theorem 3.4. Consider the quadratic function

f(x) = Ax2 + Bx + C

(1) If A is positive, then f has a smallest value. This smallest value is C − B2

4A

and is obtained for the input x = − B2A

.

3. QUADRATIC FUNCTIONS 15

(2) If A is negative, then f has a largest value. This largest value is C − B2

4Aand

is obtained for the input x = − B2A

.

Proof. To answer the question we complete the square.

f(x) = Ax2 + Bx + C = A(x + B

2A

)2+ C − B2

4A.

Suppose A > 0. Then A(x + B

2A

)2 ≥ 0 and f assumes its least value if A(x + B

2A

)2=

0. This happens for xs = − B2A

and f(xs) = C − B2

4Ais the least value of f .

Suppose A < 0. Then A(x + B

2A

)2 ≤ 0 and f assumes its largest value if A(x + B

2A

)2=

0. This happens for xl = − B2A

and f(xl) = C − B2

4Ais the largest value of f . �

Exercise 3.5. Find the largest or the smallest value, whichever there is, of the givenquadratic function and the input for which the extreme value is obtained.

(1) f1(x) = x2 + 2x(2) f2(x) = −x2 + 2x + 2(3) f3(x) = 2x2 − 3x + 2(4) f4(x) = −2x2 − 2x− 1

3

(5) f5(x) = −x2 + 3x− 12

Exercise 3.6. Work Example 3.3 if

(1) the farmer has 2000 feet of fence.(2) the farmer has 4000 feet of fence.(3) the farmer has a feet of fence.

Theorem 3.7. A cannon is being fired and the cannon ball exits the cannon with ahorizontal velocity of vx [meters/second] (= [m/s]) and a vertical velocity of vy [m/s].Then, using the theory of gravity of Isaac Newton (1642-1727), the horizontal locationx of the cannon ball and the elevation y of the cannon ball are given by the formulas

x = vxt, y = −12gt2 + vyt + y0

where

• g = 9.81 [m/s2] is the gravity constant;• t is time in seconds, and t = 0 at the moment when the ball exits the barrel;• x is the horizontal distance in meters of the cannon ball from the cannon, so

x = 0 when the bullet exists the barrel;• y is the elevation of the cannon ball in meters, and the end of the barrel is

the elevation y = y0.


y0

y

x

Example 3.8. Suppose that a cannon is being fired and the bullet exits the barrelthat is y0 = 1 [m] above ground, with the horizontal velocity vx = 10 [m/s] and thevertical velocity vy = 20 [m/s]. To simplify the computations we take 1

2g = 5 [m/s2].

Hence our equations are in this case

x = 10t, y = −5t2 + 20t + 1.

(1) In order to find the highest elevation that the cannon ball reaches we completesquares.

y = −5(t− 2)2 + 21

Therefore the cannonball reaches the height of 21 meters after 2 seconds. Wemust be dealing with a toy cannon.

(2) To see when the ball hits the ground (elevation y = 0) we must solve thequadratic equation

−5t2 + 20t + 1 = 0.

We find using the quadratic formula or (1) that

t = 2±√

21/5.

Of course, the time must be positive and hence t = 2 +√

21/5 is the correctanswer, approximately 4 seconds.

(3) How far from the cannon does the bullet hit the ground? After 4 secondsthe bullet has traveled the horizontal distance x = 10 · 2 = 20 meters.

Example 3.9. Suppose that a cannon is being fired and the bullet exits the barrelthat is y0 = 10 [m] above ground, with the horizontal velocity vx = 500 [m/s] andthe vertical velocity vy = 400 [m/s]. To simplify the computations we take 1

2g = 5

[m/s2]. Hence our equations are in this case

x = 500t, y = −5t2 + 400t + 10.

(1) In order to find the highest elevation that the cannon ball reaches we completesquares.

y = −5(t− 40)2 + 8, 010

4. THE STORY OF POLYNOMIAL EQUATIONS 17

Therefore the cannonball reaches the height of 8, 010 meters after 40 seconds.(2) To see when the ball hits the ground (elevation y = 0) we must solve the

quadratic equation

−5t2 + 400t + 10 = 0.

We find using the quadratic formula or (1) that

t = 40±√

1602.

Of course, the time must be positive and hence t = 40+√

1602 is the correctanswer, approximately 80 seconds.

(3) How far from the cannon does the bullet hit the ground? After 4 secondsthe bullet has traveled the horizontal distance x = 500 · 80 = 40, 000 meters.

It would be interesting to obtain realistic data for the problem. Any volunteers fromROTC?

Exercise 3.10. Answer the questions raised and answered in Example 3.8 and Ex-ample 3.9 for the given values. [Use (1/2)g = 5 [m/s2].]

(1) y0 = 0, vx = 400 [m/s], vy = 200 [m/s].(2) y0 = 0, vx = 440 [m/s], vy = 200 [m/s].(3) y0 = 0, vx = 300 [m/s], vy = 150 [m/s].(4) y0 = 0, vx = 300 [km/hr], vy = 150 [km/hr]. Note that here either the km/hr

(kilometers per hour) have to be converted to m/s (meters per second) orthe [m/s2] must be converted to [km/hr2].

Exercise 3.11. A ball is thrown straight up with an initial velocity of vy = 10 [m/s]from an elevation of y0 = 2 [m]. What elevation does the ball reach and when will ithit the ground? [Note that this is just the cannon ball problem with vx = 0.]

Exercise 3.12. An object is dropped from an elevation of y0 = 20 [m]. When will ithit the ground? [Note that this is just the cannon ball problem with vx = vy = 0.]

4. The story of polynomial equations

Definition 4.1. A cubic equation or an equation of degree 3 is an equationthat can be changed be simplified by algebraic manipulations to the form

(4.2) Ax3 + Bx2 + Cx + D = 0, where A 6= 0, B, C, D are given numbers.

Example 4.3.

(1) For A = 1, B = C = 0, D = 8, we get the cubic equation

x3 + 8 = 0, x = −2.

(2) For A = 3, B = 0, C = 1, D = 3, we get the cubic equation

3x3 + x + 3 = 0, x =?.


(3) The equation2x3 + 4x = −3x2 + 2x + 7

simplifies to2x3 + 3x2 + 2x− 7 = 0, x =?.

Lemma 4.4. If x is a solution of (4.2), then y = x+ B3A

is a solution of (the simpler)equation

(4.5) Ay3 +

(C − B2

3A

)y +

(2B3

27A2− BC

3A+ D

)= 0

and if y is a solution of (4.5), then x = y− B3A

is a solution of the original equation.

Example 4.6. Consider the equation

x3 + 2x2 − x− 2 = 0.

Here A = 1, B = 2, C = −1, D = −2. Hence

C − B2

3A= −7

3, 2B3

27A2 − BC3A

+ D = −2027

,

and we get the “simpler” equation

y3 − 73y − 20

27= 0.

The unknowns are related by the formula x = y − B3C

= y − 23. Now x = 1 is a

solution of the given equation (check!), so y = 53

must be a solution of the second

equation which is true. Also y = −43

is a solution of the second equation (Check!)

hence x = −43− 2

3= −2 must be a solution of the first equation which is true.

As in the case of quadratic equations we note first that the special equation x3 = ais always solvable. This has to do with the nature of the real numbers. In fact, wehave the following general theorem.

Theorem 4.7. The equation xn = a with a ≥ 0 a unique solution x ≥ 0.

Definition 4.8. The nth root of a number a ≥ 0, denoted n√

a, is the unique solution≥ 0 of the equation xn = a.

History

• Scipione del Ferro (1465-1526) solves x3 + mx = n around 1515; reveals thesecret to his student Antonio Fior.

• Niccolo Fontana (“Tartaglia”, the stutterer) (1500-1557) solves x3 + px2 = naround 1535.

• Fior challenges Tartaglia to a public contest. Tartaglia solves x3 + Bx2 +Cx + D = 0 and wins the contest.

• Girolamo Cardano (“unprincipled genius”) (1501-1576) gets the solutionfrom Tartaglia by promising to keep it a secret.


• Cardano publishes “Ars Magna” around 1545 with the solution of the cubicequation.

• Public accusations of plagiarism.• Ludovico Ferrari (1522-1565), a student of Cardano, solves the general quar-

tic equation

a4x4 + a3x

3 + a2x2 + a1x + a0 = 0.

• Niels Henrik Abel (1802-1829) shows that there is no “solution formula” forthe general quintic equation

a5x5 + a4x

4 + a3x3 + a2x

2 + a1x + a0 = 0.

• Evariste Galois (1811-1832) develops a revolutionary theory about the exis-tence of solution formulas for the general polynomial equation

anxn + an−1x

n−1 + · · ·+ a2x2 + a1x + a0 = 0.

Here is how the solution formula for the cubic

(4.9) x3 + mx = n

was derived. The following formula is easily verified.

(4.10) (a− b)3 + 3ab(a− b) = a3 − b3

Comparing (4.9) with (4.10) we conclude that if we can find a, b such that

(4.11) 3ab = m, and a3 − b3 = n

then x = a− b is a solution of (4.9). Now from (4.11) we obtain that a = m3b

and by

substitution that n = m3

27b3− b3, hence

27b6 + 27nb3 −m3 = 0

must be satisfied by b. This really is a quadratic equation for b3. In fact, it is

27(b3)2 + 27n(b3)−m3 = 0.

By the quadratic formula we have

b3 = 12·27

(−27n±

√(27n)2 + 4 · 27m3

)= −n

2± 1

18

√81n2 + 12m3.

We can now find a = m/3b and the solution x = a− b.

Example 4.12. Consider the cubic

x3 + x = 1

Then m = n = 1, and

(4.13) b3 = −1

2± 1

18

√93.

We can compute a = m/(3b) = 1/(3b) and get the solution x = a − b. In fact,using the calculator and the plus sign in (4.13), b3 ≈ 0.036, b ≈ 0.33, a ≈ 1.01, and


x = a− b ≈ 0.68. If we compute x3 + x we get x3 + x ≈ (0.68)3 + (0.68) = 0.99 ≈ 1,as it should be. Using the minus sign in (4.13), we end up with the same solution.

So far so good, but there was a shocking observation. Consider

x(x− 1)(x + 1) = x3 − x = 0

that clearly has the solutions x = 0, x = 1, x = −1. In this case m = −1 and n = 0.Hence

b3 = ± 118

√−12 = ±1

9

√−3.

This means that b3 and hence b, and hence a make no sense, yet x = 0, x = 1, andx = −1 are perfectly good solutions of x3−x = 0. This fact motivated mathematiciansto accept a larger system of complex numbers as “numbers”

C = {a + bi | a, b ∈ R, i2 = −1}.It was Carl Friedrich Gauss (1777-1855) that put complex numbers on a sure footingand, using the complex numbers, the solution formula for the cubic works perfectlywell.Gauss also proved the Fundamental Theorem of Algebra that says that everypolynomial equation

anxn + an−1x

n−1 + · · ·+ a2x2 + a1x + a0 = 0.

has solutions in the system of C of complex numbers.

Exercise 4.14. Check that the given values are solutions of the given polynomialequations.

(1) x = −2, x3 + 3x2 + 3x + 2 = 0.(2) x = 1, x4 − x3 + 2x2 + x− 3 = 0.(3) x = −1/2, x5 + 3

2x4 + 1

2x3 + x + 1

2= 0.

(4) x = 1, x = 2, x = 3, x = 4, x4 − 10x3 + 35x2 − 50x + 24 = 0.

Exercise 4.15. Compute to get a simple fraction of the form a/b, a, b integers.

(1) 512

+ 38

=

(2)3574

=

(3)34

56+ 1

4

=

(4)56+ 1

434

=

Exercise 4.16. A polynomial is an expression that can be changed to the “standardform”

anxn + an−1x

n−1 + · · ·+ a2x2 + a1x + a0,


for specific values of n, an, . . . , a0. Change the following expressions to standard formand state the specific values of n, an, . . . , a0.

(1) 2x3 + 3x2 + 4x + 5(2) 2x3 − 3x2 + 4x− 5(3) −2x3 + 3x2 − 4x + 5(4) −2x3 − 4x + 5(5) 3x4 − 2x + 3

Exercise 4.17. By multiplying out and combining terms change the following ex-pressions to the form

anxn + an−1x

n−1 + · · ·+ a2x2 + a1x + a0.

(1) 2(x + 32)2 − 4

(2) (x− 12)3

(3) (2x− 1)3

(4) (x + 2)2(x− 2)(5) (x + 3)2(x− 2)2

(6) (x− (1−√

5))(x− (1 +√

5))

Exercise 4.18. Equalities such as

(1) (a + b)2 = a2 + 2ab + b2.(2) (a + b)3 = a3 + 3a2b + 3ab2 + b3.(3) (a− b)(a + b) = a2 − b2.(4) (a− b)(a2 + ab + b2) = a3 − b3.

are identities because they remain true if the letters are replaced by an other mean-ingful arithmetic term. E.g. replacing a by x and b by 2 in (2) results in the identity(x + 2)3 = x3 + 6x2 + 12x + 8 valid for any value of x. For the following identi-ties state from which one of the above identities they are derived by making whichsubstitutions.

(1) (n + 3)(n− 3) = n2 − 9.(2) (2y + 1)3 = 8y3 + 12y2 + 6y + 1.(3) (a− 4)(a + 4) = a2 − 16.(4) x3 − 27 = (x− 3)(x2 + 3x + 9)(5) x3 + 27 = (x + 3)(x2 − 3x + 9)(6) (3− p)2 = 9− 6p + p2.

Exercise 4.19. It can be disproved that an equation is an identity by finding valuesfor which the supposed identity becomes false. E.g. a(b−c) = ab−c is not an identitybecause for a = 2, b = 2, c = 3 (randomly chosen values, but using 0 and 1 shouldavoided because these two numbers behave in a peculiar way), the left hand side ofthe supposed identity is −2 and the value of the right hand side is 1. Decide whetherthe equalities below are identities in which case you should have a justification, andif not, make an example for which they fail.


(1) (u + v)2 = u2 + v2.

(2)√

a2 + 4 = a + 2.

(3) (a + b)(c + d + e) = ac + ad + ae + bc + bd + be.

(4)(a + b)c

(d + e)c=

a + b

d + e.

(5)a + bc

d + ec=

a + b

d + e.

(6)a + b

c + d=

a

c+

b

d.

Exercise 4.20. In the following an equation and a value are given. Decide whetheror not the given value solves the given equation.

(1) n4 + 3n2 + 1 = 0, n = −2.(2) (x− 2)(x + 5)(x− 2) = 0, x = 3.(3) (y + 1)(y2 + y + 1) = 0, y = −1.(4) (y + 1)(y2 + y + 1) = 0, y = 1.(5) n4 − 3n3 − 6n2 + 8n = 0, n = 0.(6) n4 − 3n3 − 6n2 + 8n = 0, n = 4.(7) t2 − 2t− 1 = 0, t = 1 +

√2.

5. Solutions of Exercises of Chapter I

1.1. Let A1 be the area of the field yielding 2/3 of a bushel per square yard, and letA2 be the area of the field yielding 1/2 of a bushel per square yard. Then A1 = 1200square yard, and A2 = 600 square yard.1.2. Let B1 be the measures in the bundles of type 1, B2 be the measures in thebundles of type 2, and B3 be the measures in the bundles of type 3. There are threeconditions that must be satisfied:

3B1 + 2B2 + B3 = 39, 2B1 + 3B2 + B3 = 34, B1 + 2B2 + 3B3 = 26.

The solutions are B1 = 374,B2 = 17

4,B3 = 11

4.

1.3 For example, pick 7. Then 7 + 237 = 35

3, 35

3− 1

3353

= 709, and finally 70

9− 1

10709

= 7.To show that this always works, replace in the above 7 by an arbitrary value x:x + 2

3x = 5

3x, 5

3x− 1

353x = 10

9x, and finally 10

9x− 1

10109x = x.

1.4 Answer: 138697

.1.5 Let x1 ≤ x2 ≤ x3 ≤ x4 ≤ x5 be the shares. If x1 = x2 = s (for small), andx3 = x4 = x5 = ` (for large), then s = 25/4 and ` = 175/6. If x1 = a and thedifference between consecutive x’s is d, then a = 5/3 and d = 55/6.1.12 x = 6/13.

5. SOLUTIONS OF EXERCISES OF CHAPTER I 23

1.13 x = −3/17.1.14 x = 6.1.15 x = 0.1.16 x = 1.1.17 x = 8/5.1.18 No solution.1.19 Every number x is a solution.1.25 (1(a)) V = 5632/21; (1(b)) r = 3

√63/352; (2(a)) b = 6; (2(b)) h = 6; (2(c))

V = 50.1.26 (1) F = 95; (2) C = 40; (3) F = 80/7 or C = −80/7; (4) C = 80/3 or F = 80;(5) C = 160 or F = 320; (6) C = −320/13 or F = −160/13.1.33 (1) x = −4/11, y = 5/11; (2) x = 6/5, y = 14/5; (3) x = 2, y = 4; (4) x = 7/17,y = −13/34; (5) no solution; (6) infinitely many solutions, for any y, y together withx = 1 − 3y is a solution. E.g. y = 0 and x = 1 is a solution, y = 1 and x = −2 is asolution.1.34 Let g, m, b be the prices of good, mediocre, and bad crop, respectively. Theequations to solve are

3g + 2m + b = 39

2g + 3m + b = 34

g + 2m + 3b = 26

Then g = 37/4, m = 17/4, b = 11/4 are the unique solutions.1.35 Let h be the value of a horse. Then h = 100.1.36 The father is 35, the son 5.1.37 Ann is 15.1.38 5/3 grams must be added.1.39 There were initially 18 males and 12 ladies.2.9

(1) x2 + 5x =(x + 5

2

)2 − 254.

(2) 2x2 + 5x = 2(x + 5

4

)2 − 258.

(3) 2x2 + 5x + 3 = 2(x + 5

4

)2 − 18.

2.10

(1) x2 − 5x =(x− 5

2

)2 − 254.

(2) 3x2 − 5x = 3(x− 5

6

)2 − 2512

.

(3) 3x2 − 5x + 4 = 3(x− 5

6

)2+ 23

12.

2.11

(1) x2 − 23x =

(x− 1

3

)2 − 19.

(2) 3x2 − 2x = 3(x− 1

3

)2 − 13.

(3) 3x2 − 2x + 4 = 3(x− 1

3

)2+ 11

3.


2.12

(1) x2 − 23x =

(x− 1

3

)2 − 19.

(2) 12x2 − 3

4x = 1

2

(x− 3

4

)2 − 932

.

(3) 12x2 − 3

4x + 3

2= 1

2

(x− 3

4

)2+ 39

32.

2.17 The solutions are

(1) x = 12

(−1±

√5).

(2) x = 0 or x = 32.

(3) x = 14

(−1±

√57

).

(4) x = 12

(−1±

√−39

), no solution.

(5) x = 13

(2±

√19

).

(6) No solution.(7) x = ±

√3.

(8) x = −2±√

6.(9) x = 1±

√6

3.5

(1) f1 assumes its smallest value of −1 at x = −1.(2) f2 assumes its largest value of 3 at x = 1.(3) f3 assumes its smallest value of 7/8 at x = 3/4.(4) f4 assumes its largest value of 1/6 at x = −1/2.(5) f5 assumes its largest value of 7/4 at x = 3/2.

3.6

(1) The largest area is 500,000 square feet for x = 500 feet and y = 1000 feet.(2) The largest area is 2,000,000 square feet for x = 1000 feet and y = 2000 feet.

(3) A = −2(x− a

4

)2+ a2

8. Hence the largest area of a2

8square feet is obtained

for x = a4

feet and y = a2

feet.

3.10

(1) The cannon ball rises to 2000 meters after 20 seconds and hits ground at16, 000 meters or 16 kilometers.

(2) The cannon ball rises to 2000 meters after 20 seconds and hits ground at17, 600 meters or 17.6 kilometers.

(3) The cannon ball rises to 1125 meters after 15 seconds and hits ground at9, 000 meters or 9 kilometers.

(4) 300 km/hr = 30010003600

m/s = 2503

m/s ; 150 km/hr = 15010003600

m/s =1253

m/s The cannon ball rises to 3125/36 meters after 25/6 seconds andhits ground at 3125/9 meters.

3.11 The ball reaches 7 meters after 1 second and hit ground after 1 + 15

√35 ≈ 2.18

seconds.3.12 The object hits ground after 2 seconds.

5. SOLUTIONS OF EXERCISES OF CHAPTER I 25

4.14 Substitute the given value on the left and compute. You must get 0.4.15 (1) 19/24; (2) 12/35; (3) 9/13; (4) 13/9.4.16 (1) n = 3, a3 = 2, a2 = 3, a1 = 4, a0 = 5; (2) n = 3, a3 = 2, a2 = −3, a1 = 4,a0 = −5; (3) n = 3, a3 = −2, a2 = 3, a1 = −4, a0 = 5; (4) n = 3, a3 = −2, a2 = 0,a1 = −4, a0 = 5; (5) n = 4, a4 = 3, a3 = 0, a2 = 0, a1 = −2, a0 = 3.4.17 (1) 2x2+6x+ 1

2; (2) x3− 3

2x2+ 3

4x− 1

8; (3) 8x3−12x2+6x−1; (4) x3+2x2−4x−8;

(5) x4 + 2x3 + 13x2 − 12x + 36.4.18 (1): (3) with a = n, b = −3; (2): (2) with a = 2y, b = 1; (3): (3) with a = a,b = 4; (4): (4) with a = x, b = 3; (5) (4) with a = x, b = −3; (6): (1) with a = 3,b = −p.4.19 (1) False, try u = v = 1; (2) False, try a = 1; (3) True, law for multiplying out;(4) True, correct cancellation; (5) False, try a = 1, b = c = d = e = 2; (6) False, trya = b = c = d = 1. Note: There are many choices of variables that can be usedto show that an equality is not an identity. However, if the equality holds for somechoice of variables, then it is not shown that the equality is an identity. It may failfor other choices.4.20 no,no,yes,no,yes, yes, yes.

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