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Monday, April 25, 2022

A new exam to cover a topic "System of General Inequalities" has been released to UNIZOR.COM - Math 4 Teens - Algebra. It contains six multiple choice problems.
Highly recommended for those who seek Knowledge.

1. What is a solution to the following inequality? (x^{2} + y^{2} −16)·(x^{2} + y^{2} − 36) < 0

2. What is the area S of the part of a coordinate plane defined by the following inequality? (x^{2} + y^{2} − 25)·(x^{2} + y^{2} − 49) > 0

3. What is the area S of the part of a coordinate plane defined by the following system of inequalities? x^{2} + y^{2} − 4 < 0
y − x < 0

4. What is the perimeter P of the part of a coordinate plane defined by the following system of inequalities? x^{2} − 4 < 0
y^{2} − 16 < 0

5. What is the perimeter P of the part of a coordinate plane defined by the following system of inequalities? x^{2} + y^{2} − 9 < 0
x + y < 0

6. The part of a coordinate plane is defined by the following inequalities? x^{2} + y^{2} − R^{2} < 0
y − |x| < 0
What is the value of parameter R, if the area of thus defined part of a coordinated plane equals to 48π?

PROBLEM
Let X_{0} = 0, X_{1} = 1
and general formula for this sequence is X_{n+2} = 3·X_{n+1} − 2·X_{n}
Prove that Y_{n} = X_{n}^{2} + 2^{(n+2)} are exact squares of some natural numbers.
PROOF
Let's prove by induction that X_{n} = 2^{n} − 1
(a) For n=0 this is true since x_{0} = 2^{0} − 1 = 0, as set above
(b) For n=1 this is true since x_{1} = 2^{1} − 1 = 1, as set above
(c) Assume that X_{k} = 2^{k} − 1 and X_{k+1} = 2^{k+1} − 1
and, based on that, calculate X_{k+2}: X_{k+2} = 3·X_{k+1} − 2·X_{k} =
= 3·(2^{k+1} − 1) − 2·(2^{k} − 1) =
= 3·2^{k+1} − 3 − 2·2^{k} + 2 =
= 3·2^{k+1} − 2^{k+1} − 1 =
= 2·2^{k+1} − 1 =
= 2^{k+2} − 1
which corresponds to a formula that we are trying to prove.

Using this general formula for X_{k} = 2^{k} − 1, we will prove that Y_{n} = X_{n}^{2} + 2^{(n+2)} are exact squares of some natural numbers.
Indeed, Y_{n} = (2^{n} − 1)^{2} + 2^{(n+2)} =
= 2^{2n} − 2·2^{n} + 1 + 2^{(n+2)} =
= 2^{2n} − 2·2^{n} + 1 + 4·2^{n} =
= 2^{2n} + 2·2^{n} + 1 =
= (2^{n} + 1)^{2}

4^{x} + 6^{x} = 9^{x} (2·2)^{x} + (2·3)^{x} = (3·3)^{x}
Since (2·3)^{x} never equals to zero, we can divide both sides of this equation by it without losing any solutions.
The result is (2/3)^{x} + 1 = (3/2)^{x}
Let y=(2/3)^{x}
Then
y + 1 = 1/y
Since y is always positive, we can multiply both sides by it getting an equivalent equation y² + y − 1 = 0
Solutions to this quadratic equation are y_{1,2} = (−1 ± √5)/2
Therefore,
x_{1,2} = log_{2/3}[(−1 ± √5)/2]

Add two altitudes towards the bases of both triangles (big and small), one H from the top of the big triangle onto a side of 8+4=12 length and another h from the top of the small triangle onto a side of 8 length.
Let the unknown area of a small triangle be x.
Then
(a) 12·H = x + 36
(b) 8·h = x
(c) H/h = (8+3)/3
This is a simple system of three equations with three unknowns.
Divide the first equation (a) by the second (b): (12·H)/(8·h) = (x + 36)/x
or
(d) (3/2)·(H/h) = 1 + 36/x
Using the third equation (c) that states that H/h = 11/3, express (d) as (3/2)·(11/3) = 1 + 36/x
or 11/2 = 1 + 36/x
or 9/2 = 36/x
from which x = 8

Solution A
From the third equation: z = 20/y
Substitute it to the second, getting a system of two equations with two unknowns: x·y = −12 x·20/y = −15
From the second equation: y = −x·20/15 = −x·4/3
Substitute into the first equation: −x·x·4/3 = −12
Solving it for x: x² = 9
One root is x_{1}=3, another is x_{2}=−3. Casex_{1}=3: y_{1} = −4 z_{1} = −5 Casex_{2}=−3: y_{2} = 4 z_{2} = 5

Solution B (more clever)
Obviously, none of the unknowns equals to zero.
Multiply all three equations, getting (x·y·z)² = 3600
Therefore, we have two cases: x·y·z = 60
or x·y·z = −60 Casex·y·z = 60
Divide it by each given equation, getting z_{1} = −60/12 = −5 y_{1} = −60/15 = −4 x_{1} = 60/20 = 3 Casex·y·z = −60
Divide it by each given equation, getting z_{2} = 60/12 = 5 y_{2} = 60/15 = 4 x_{2} = −60/20 = −3

Answer
There are two solutions to this system: (x_{1},y_{1},z_{1})=(3,−4,−5) (x_{2},y_{2},z_{2})=(−3,4,5)

Checking
ALWAYS PERFORM CHECKING OF ALL SOLUTIONS

Example 2 (x+2)·(y+2) = −10 (x+y)·(x·y−3) = 15

Solution
Both equations are of the second degree, so directly resolving one of them for, say, y in terms of x and substituting into another might be too complex.
Notice the symmetry between unknown variables x and y.
If, instead of x, we use y and, instead of y, use x, we will have the same equation.
In cases like this it might be useful to introduce new variables a=x+y and b=x·y.
Transform these equations: (x+2)·(y+2) =
= x·y+2(x+y)+4 = −10 (x+y)·(x·y−3) =
= (x+y)·(x·y)−3(x+y) = 15
We can assign two new unknown variables a=x+y and b=x·y to obtain the following quadratic system of two equation with one of them being of the first degree b + 2a + 4 = −10 a·b − 3a = 15
From the first linear equation we can get b in terms of a and substitute into a second equation getting a quadratic equation with one variable: b = −2a − 14 a·(−2a−14) − 3a = 15
Simplifying the above: 2a² + 17a + 15 = 0
This quadratic equation has roots a_{1} = −1 a_{2} = −15/2
Each of these solutions should be considered separately to get x and y.

Casea = −1. b = −2a−14 = −12
This leads us to a quadratic system x + y = −1 x·y = −12
Expressing y in terms of x from the first equations and substituting into the second: y = −1 − x x·(−1− x) = −12 x² + x − 12 = 0 x_{1} = −4; x_{2} = 3 y_{1} = 3; y_{2} = −4
(notice the symmetry, mentioned in the beginning)

Casea = −15/2. b = −2a−14 = 1
This leads us to a quadratic system x + y = −15/2 x·y = 1
Expressing y in terms of x from the first equations and substituting into the second: y = −15/2 − x x·(−15/2 − x) = 1 x² + (15/2)·x + 1 = 0 2x² + 15x + 2 = 0 x_{3} = (−15+√209)/4;
x_{4} = (−15−√209)/4 y_{3} = (−15−√209)/4;
y_{4} = (−15+√209)/4
(notice the symmetry, mentioned in the beginning)

Answer
There are 4 (x,y) pairs of solutions: −4, 3
3, −4
(−15+√209)/4, (−15−√209)/4
(−15−√209)/4, (−15+√209)/4

Checking
ALWAYS PERFORM CHECKING OF ALL SOLUTIONS

Example 3 x·y² − x − 2y = −2 x·y + y = 2

Solution
The first equation can be transformed to an equivalent x·(y²−1) − 2(y−1) = 0 x·(y−1)·(y+1) − 2(y−1) = 0 (y−1)·(x·y+x−2) = 0
This allows to split a problem in two: Casey−1=0, that is y=1
From the second equation of the system x·y+y=2
we obtain x for y=1: x·1 + 1 = 2 x = 1
So, the first solution to our system is (x_{1}=1, y_{1}=1) Casex·y+x−2=0
Let's subtract this from the second equation x·y+y=2
The result is y − x = 0, that is y = x
which converts the second equation into a quadratic one x² + x − 2 = 0
Its roots are: x_{2} = −2
from which y_{2}=x_{2}=−2 and x_{3} = 1
from which y_{3}=x_{3}=1
which we already obtained in the previous case.

Solution
First simplification can be achieved by introducing two new variables: u = x+y v = x−y
This leads to a new simpler system of equations: 1/u + 1/v = 36/5 u² + v² = 13/18
The first on can be rewritten as: (u+v)/(u·v) = 36/5
The second one can be rewritten as: u² + 2u·v + v² − 2u·v = 13/18
or (u+v)² − 2u·v = 13/18
Another pair of variables can simplify it even further: p = u+v q = u·v
In terms of these variables the system looks like p/q = 36/5 p² − 2q = 13/18
Now we can express q as a function of p and substitute it the second equation: q = 5·p/36 p² − 2·5·p/36 − 13/18 = 0
So, we have a plain quadratic equation for p: p² − (5/18)p − 13/18 = 0
It has 2 roots: p_{1}=1 (easy to guess) and p_{2}=−13/18 (because the product of these roots must be equal to a free member of the equation).
Correspondingly, q_{1} = 5/36 q_{2} = −(13/18)·(5/36) = −65/648
Since we have defined p=u+v and q=u·v, values u and v must be the roots of the quadratic equation
w² −p·w + q = 0 Case(p,q)=(1,5/36) w² − w + 5/36 = 0
Roots are: w_{1}=1/6, w_{2}=5/6
Therefore, we have two cases: (u,v)=(5/6, 1/6) and (u,v)=(1/6, 5/6)
In the first case we can have a system of equations for x and y as x+y = u = 5/6 x−y = v = 1/6
Solution to this linear system is: x = 1/2, y = 1/3
In the second case we can have a system of equations for x and y as x+y = u = 1/6 x−y = v = 5/6
Solution to this linear system is: x = 1/2, y = −1/3 Case(p,q)=(−13/18,−65/648) w² + (13/18)·w − 65/648 = 0
Roots of this equation, w_{1} and w_{2} will give values for u and v: (u,v)=(w_{1}, w_{2}) and (u,v)=(w_{2}, w_{1})
From these two pairs of u and u we will derive x and y.
Getting exact values in this case we leave to those who are not afraid of cumbersome calculations.

Checking
JUST DON'T FORGET TO CHECK YOUR SOLUTION.

As we know, we can use decimal notation to represent real numbers. Integer numbers are represented in a decimal form without decimal point. Rational non-integer numbers use finite or infinite periodical sequence of decimal digits after the decimal point to represent their fractional part. Irrational numbers in decimal form have infinite sequence of non-periodical decimal digits after the decimal point.

In this lecture we will talk about how to use approximation of those real numbers, whose exact decimal representation is impossible (because they have an infinite sequence of decimal digits after the decimal point) or inconvenient (for lengthy sequences of digits) to show.

As a simple example, consider a task of measuring 1/7^{th} of 1 meter of a rope.
The metric ruler has centimeters and millimeters on it. It means, it's impossible to measure exactly 1/7^{th} of 1 meter because 1/7 = 0.(142857) - an infinite periodical fraction.
The best we can do is to measure 14 cm and either 2 mm or 3 mm because 0.142 < 1/7 < 0.143.
It's better to choose 14 cm and 3 mm
because the exact value 1/7 = 0.(142857) is closer to it.

The need for approximation in the above example is dictated by our practical ability and limitations to deal with exact number. These limitations determined the level of precision required from approximation - 1 millimeter.

Consider another example - a population of the United States of America.
Obviously, we cannot know the exact number of people living in the country because people get born, die, move to another country or come from another country all the time.
So, for practical reasons we approximate this number to, say, millions. If we say that the population of the USA is, approximately, 330 millions, it's sufficient to address most issues related to the population, like how much water is consumed by all people in a country or average density of the population.
Again, practical considerations dictate certain level of approximation, its precision - 1 million of people.

Now we approach the approximation more formally.
The purpose of the approximation is to represent the real number in question by another number, according to certain rules:

1. Out of all real numbers we choose a countable subset of numbers equally spaced from each other, we will call them base numbers.
For example, we can have a set of all integer numbers or a set of numbers, starting at zero, with a distance of 1000 from each other, or a set of numbers, starting at zero, with a distance of 0.001 from each other.
In the example above, when we measured the length with a precision of 1 millimeter = 0.001 meter, our base numbers are 0, 0.001, 0.002, 0.003 etc.
In the example above, when we counted people in millions, our base numbers are 0, 1000000, 2000000, 3000000 etc.
This set of base numbers determines the possible approximate values for numbers we would like to represent.

2. For any real number we would like to represent approximately we use the closest to it number from a set of base numbers chosen above.

3. A special case, when the exact number is equidistant from both base numbers, the one less and the one greater than it, needs special consideration that we discuss below.

Example 1
We need to find a length of a side of a square whose area is 2 m².
From geometry we know that this side should be equal to a square root of 2 meters, but it's an irrational number, so we cannot represent it exactly in decimal notation needed for practical purposes.
We choose the precision sufficient for our practical purposes and practically achievable using the tools at hand (say, a measuring tape with meters and centimeters), to be 1 centimeter, that is 0.01 of a meter.
This determines the base numbers for our approximation 0, 0.01, 0.02, 0.03 etc.
Then we approximate square root of 2 beyond the second digit after the decimal point to determine which base number it's closest to: √2 ≅ 1.414
which is closer to a base number 1.41 than to 1.42.
So, with a precision of 0.01 (that is, in centimeters) the size of a side of a square with area of 2 square meters equals to 1.41 m

Example 2
The distance between two cities is represented in kilometers. The exact distance is usually measured between the main post offices in these cities, and it's never equals to an exact integer number of kilometers.
For example, the distance from Hanoi to Shanghai is listed as 1925 km.
Obviously, it's an approximation to an integer number of kilometers.
The actual exact distance from the entrance door to Hanoi's main post office to the entrance door to Shanghai's main post office along the shortest route might be something like 1924.532 km (that is, 1924 km and 532 m), but we replaced the exact distance with its approximation as a more practical value.

Special case

It's logical and reasonable to approximate any exact number with a base number that is closest to it.

Consider now a case when an exact number is equidistant from two base numbers, one below and another above it.
For example, we would like to approximate to a precision of 1000 (so, base numbers are 0, 1000, 2000, 3000 etc.) and a number we want to approximate is 56500, which is equidistant from base numbers 56000 and 57000.

Or we would like to approximate to a precision of 0.01 (so, base numbers are 0, 0.01, 0.02, 0.03 etc.) and a number we want to approximate is 8.565, which is equidistant from base numbers 8.56 and 8.57.

This is a dilemma that must be resolved by some rule imposed on the process of approximation.
Unfortunately, there are more than one rule, and these rules might contradict each other. Fortunately, one rule does play a dominant role and used in most cases. This is the one we would like to specify as the one to follow, unless a specific other rule is mentioned. In case an exact number is equidistant from two base numbers, one below and one above it, choose the one that has higher absolute value.

Examples:
Precision is 0.001,
exact value is 25.6575,
approximate value is 25.658.
Precision is 0.001,
exact value is −2.7595,
approximate value is −2.760.
Precision is 10,
exact value is 25,
approximate value is 30.
Precision is 10,
exact value is −95,
approximate value is −100.

From Approximation to Exact

Let's reverse the logic and try to evaluate the exact number, if it's approximation is given.
Again, a lot depends on the precision of approximation. The more precise approximation was applied to exact number - the closer the approximation is to it and, therefore, more precise evaluation of the exact number can be performed, if an approximate number is known.

Let's assume, we have an exact number X and its approximate value with precisionδ is R. It means that among base values R−δ, R and R+δ the number X is closer to R than to R−δ or R+δ.

If so, the following inequality must hold R−δ/2 ≤ X < R+δ/2
as illustrated below with the blue area representing the possible values of X, if R is its approximation with precision δ.
So, if approximate value of the number of people living in New York in 2019 was 8.419 million (implying the precision δ=0.001 of a million, that is δ=1000), the exact number was greater or equal to 8.4185 million (that is, 8,418,500), but less than 8.4195 million (that is, 8,419,500).

If approximate value for number π is 3.14 (implying the precision δ=0.01), that exact ratio of a circumference of any circle to its diameter is greater or equal to 3.135, but less than 3.145.

As you see, the range of values for an unknown exact number X around its known approximation R is equal to a precision δ of approximation: (R+δ/2) − (R−δ/2) = δ

Approximation Error

What happens, when we make some arithmetic operation with approximate values?
Unfortunately, the error of approximation is accumulated and growing.
Let unknown exact number X be approximated by a known number R with precision δ. Let unknown exact number Y - by a known number S with the same precision δ.
Let's determine the range of values for X+Y.

According to the laws of approximation, R−δ/2 ≤ X < R+δ/2 S−δ/2 ≤ Y < S+δ/2
Then, adding these inequalities, we obtain R+S−δ ≤ X+Y < R+S+δ
The range of values for X+Y is (R+S+δ) − (R+S−δ) = 2δ
As you see, the precision of evaluating the sum of two exact values, using the sum of their approximations, is now 2δ. So, the quality of approximation is worsening, when we do arithmetic operations with approximate values to evaluate the result of these operations with unknown exact values.

Obviously, similar worsening of the quality of approximation is observed with any other operation (like subtraction, multiplication, division etc.), when we use approximate values in order to evaluate the result of this operation on unknown exact values.

For example, we need to evaluate the rise of the sea level in 10 years from now.
According to EPA, if sea level in 1880 is taken as level zero, by 1994 this level was 6.307" = 160.192 mm above zero.

From 1994 to 2019 the sea level was rising by an average factor of μ = 1.016 a year with standard deviation σ = 0.012, reaching the level of 240.775 mm
Let's disregard extreme values and limit ourselves to 68% probability of the exact value of the average factor of the rise of a sea level to be within [μ−σ;μ+σ] range, that is from 1.004 to 1.028.

Then in 10 years from 2019 the minimum sea level will be W_{min} = 240.775·1.004^{10} ≅ 251 (millimeters above 1880 level)
while its maximum will be W_{min} = 240.775·1.028^{10} ≅ 317 (millimeters above 1880 level)
Which numbers to believe?

What is the estimate of the sea level in 100 years? W_{min} = 240.775·1.004^{100} ≅
≅ 359 mm = 0.359 meters above 1880 level W_{min} = 240.775·1.028^{100} ≅
≅ 3810 mm = 3.81 meters above 1880 level
As you see, the difference is dramatic! This proves how skeptical we have to be when hearing different predictions about future.

The formula for solutions of any quadratic equation is known.
For an equation A·X^{2}+B·X+C=0
the solutions are:

X_{1,2} =

−B±√B²−4AC

2A

There is a formula for a general cubic equation (Cardano formula), but it's very complex, and we will not list it here.

So, what can someone do to solve a cubic equation?
In some cases there is a possibility to guess one of the solutions. Then, using the polynomial division, we can reduce our cubic equation with one guessed solution to a quadratic one to find two other solutions.

Consider a cubic equation X^{3} + 4X^{2} −11X −30 = 0
As we know, any polynomial of the third order, like the one above, has three (generally speaking, complex) roots, that is values of an unknown that cause the value of the polynomial to be equal to zero. Assume these roots are X_{1} = a X_{2} = b X_{3} = c
Then, according to a corollary to a Fundamental Theorem of Algebra, our polynomial can be represented as X^{3} + 4X^{2} −11X −30 =
= (X−a)·(X−b)·(X−c)

Then, knowing one root (say, X_{1} = a) we can construct the result of multiplication (X−b)·(X−c) by dividing our original polynomial by (X−a).
This process is similar to long division of numbers.

Notice that, if we multiply (X−a)·(X−b)·(X−c), the only element without an unknown will be −a·b·c.
Therefore, −30 = −a·b·c.

If the free member of our equation is integer (and it is, it's equal to −30), it's reasonable to attempt, firstly, to find an integer root of our equation among its divisors.

Number 30 has only three divisors: 2, 3 and 5. Let's check if one of them (say, 5) is the root of a given polynomial. 5^{3} + 4·5^{2} −11·5 −30 =
= 125 + 100 −55 − 11 = 40
So, X=5 is not a root.

How about X=−5? (−5)^{3} + 4·(−5)^{2} −11·(−5) −30 =
= −125 + 100 + 55 − 30 = 0
Lucky guess! We found a root a=−5 and now we can divide the original polynomial of the third order by X−a = X−(−5) = X+5
obtaining the polynomial of the second order that should have two other roots b and c.

Here is this division, step by step.
Step 1:
Dividend: X^{3}+4X^{2}−11X−30
Divisor: X+5
Quotient: X^{2}
Multiply: (X+5)·X^{2}=X^{3}+5X^{2}
Remainder: (X^{3}+4X^{2}−11X−30)−(X^{3}+5X^{2})=
= −X^{2}−11X−30

Combine all quotients and get:
Dividend: X^{3}+4X^{2}−11X−30
Divisor: X+5
Quotient: X^{2}−X−6
Multiply: (X+5)·(X^{2}−X−6) =
= X^{3}+4X^{2}−11X−30
Remainder: zero

Now we can either attempt to find a root of the equation X^{2}−X−6 = 0
and perform a polynomial division or, considering this is an equation of the second order, just use the formula for its roots.

The guess and division approach is more educational, so we will choose it.
The free member of the polynomial we deal with now is −6, so we will try its divisors 2 or 3.
Start with a root b=2. 2^{2}−2−6 = −4
So, 2 is not a root.
How about b=−2? (−2)^{2}−(−2)−6 = 4+2−6 = 0
Lucky guess! We found a second root b=−2 and now we can divide the original polynomial X^{2}−X−6 of the second order by X−b = X − (−2) = X + 2

Let's do a division to reduce our second order polynomial X^{2}−X−6 as a product of (X+2) and a first order polynomial.
Step 1:
Dividend: X^{2}−X−6
Divisor: X+2
Quotient: X
Multiply: (X+2)·X = X^{2}+2X
Remainder: (X^{2}−X−6)−(X^{2}+2X)=
= −3X−6

Combine all quotients and get:
Dividend: X^{2}−X−6
Divisor: X+2
Quotient: X−3
Multiply: (X+2)·(X−3) =
= X^{2}−X−6
Remainder: zero

By guessing and using the polynomial divisions we have two roots (X=−5 and X=−2) and representation of the original equation as X^{3}+4X^{2}−11X−30 =
= (X+5)·(X+2)·(X−3)
The last multiplier reveals the third root of the original cubic equation X=3

CONCLUSION
Knowing one root X=a of a polynomial of the Nth order P^{(N)}(X), we can represent this polynomial as a product of (X−a) and a polynomial of the (N−1)th order P^{(N−1)}(X), thus simplifying a task of finding all the roots.