*Notes to a video lecture on http://www.unizor.com*

__Algebra+ 05__

*Problem A*

Given a system of two equations with three unknown variables

*,*

**x***and*

**y***:*

**z**

**x + y + z = A**

**x**^{−1}+ y^{−1}+ z^{−1}= A^{−1}Prove that one of the unknown variables equals to

*.*

**A***Hint A*

System of equations

**x + y = p**

**x · y = q**fully defines a pair of numbers (generally speaking,

*complex*numbers) as solutions to a quadratic equation

*.*

**X² − p·X + q**Indeed, if

*and*

**X**_{1}*are the solution of the equation, then, according to the Vieta's Theorem,*

**X**_{2}*and*

**X**_{1}+ X_{2}= −(−p) = p

**X**_{1}· X_{2}= q(See a lecture

*Math 4 Teens - Algebra - Quadratic Equations - Lecture*on UNIZOR.COM)

The only unresolved issue is:

which unknown variable takes which value from a pair

is it (

*) or (*

**x=X**_{1},y=X_{2}*).*

**x=X**_{2},y=X_{1}From this follows that, if the following system of equations is given

**x + y = a + b**

**x · y = a · b**then either (

*)*

**x=a,y=b**or (

*).*

**x=b,y=a***Proof A*

**x + y = A − z**

**x**^{−1}+ y^{−1}= A^{−1}− z^{−1}None of the unknown variables can be equal to zero, since each is represented in the second equation in the denominator.

Therefore, we can multiply the second equation by

*getting*

**x·y**

**y + x = x·y/A − x·y/z**Using the first equation, substitute

*into the second getting a system of equations*

**x+y**

**x + y = A − z**

**A − z = x·y·(1/A − 1/z)**or

**x + y = A − z**

**x·y = (A−z)/(1/A − 1/z)**or

**x + y = A − z**

**x·y = −A·z**or

**x + y = A + (−z)**

**x·y = A·(−z)**Therefore, either

**x=A, y=−z**or

*.*

**y=A, x=−z***Problem B*

Prove that

**(x + y)**^{4}≤ 8·x^{4}+ 8·y^{4}*Proof B*

Let's start with analysis of this problem.

Assume, this inequality (call it "statement

*") is true and make*

**A***invariant*(

*reversible*and

*equivalent*) transformations to it, trying to get to an obviously

**true**statement

*.*

**B**Then, using the fact that our transformations were invariant, we can say that we can start with obviously

**true**statement

*and, using the reverse transformations, derive statement*

**B***, that is we will prove that*

**A***is*

**A****true**.

Notice that, if we divide both sides of this inequality by

*and assign*

**y**^{4}*, we will reduce the number of variables from two to one, which seems to simplify the task.*

**t=x/y**Dividing by positive

*is an invariant transformation of an inequality, except a case of*

**y**^{4}*. The case of*

**y=0***can be considered separately, and in this case the inequality is obviously true since*

**y=0***.*

**x**^{4}≤ 8·x^{4}After dividing by

*and substituting*

**y**^{4}*the new inequality looks like*

**t=x/y**

**(t + 1)**^{4}≤ 8t^{4}+ 8which seems to be simpler to prove.

Let's open the parenthesis and bring all items to one side of an inequality - obviously invariant transformation

**7t**^{4}− 4t^{3}− 6t^{2}− 4t + 7 ≥ 0Notice that the sum of coefficients of a polynomial on the left is zero. That means that

*is a root of this polynomial, that is it's equal to zero for*

**t=1***.*

**t=1**Recall the

*Fundamental Theorem of Algebra*(see

*Math 4 Teens*course on UNIZOR.COM, menu items

*Algebra*-

*Fundamental Theorem of Algebra*and its

*Corollary 1*) that states that if

*is a root of a polynomial*

**x=a***of*

**P**^{(n)}**(x)***n*

^{th}degree, then this polynomial is divisible by

*, that is*

**x−a**

**P**^{(n)}**(x) = (x−a)·Q**^{(n−1)}**(x)**where

*is a polynomial of a degree lower by 1 than*

**Q**^{(n−1)}**(x)***.*

**P**^{(n)}**(x)**Therefore, since

*is a root of the polynomial of the 4th degree above, we can represent that polynomial as*

**t=1***multiplied by another polynomial of the 3rd degree.*

**(t−1)**

**7t**

= (t − 1)·(7t^{4}− 4t^{3}− 6t^{2}− 4t + 7 == (t − 1)·(7t

^{3}+ 3t^{2}− 3t − 7)Consider the polynomial

**7t**^{3}+ 3t^{2}− 3t − 7The sum of its coefficient is zero too. Therefore, we can represent it as a product of

*and a polynomial of the second degree*

**(t−1)**

**7t**

= (t − 1)·(7t^{3}+ 3t^{2}− 3t − 7 == (t − 1)·(7t

^{2}+ 10t + 7)So, the inequality we have to prove was transformed into this one:

**(t − 1)**^{2}·(7t^{2}+ 10t + 7) ≥ 0In this inequality the member

*is always greater or equal to zero.*

**(t−1)**^{2}Quadratic polynomial

*has*

**7t**^{2}+10t+7*discriminant*Δ

*, which is negative and, consequently, it has no roots, it's always not equal to zero.*

**=10**^{2}−4·7·7=−96It can only be greater than zero since the coefficient at

*is positive.*

**t**^{2}Therefore, this polynomial is always greater than zero.

That concludes the analysis of our problem.

The

**proof**proper is to start from an obviously truthful statement

**(t − 1)**^{2}·(7t^{2}+ 10t + 7) ≥ 0and transform it into

**(t + 1)**^{4}≤ 8t^{4}+ 8Replacing

*with*

**t***(recall, a trivial case*

**x/y***was already checked, so now we assume that*

**y=0***) and multiplying by*

**y≠0***finishes the proof.*

**y**^{4}*Problem C*

Prove the following inequality

**x**^{12}− x^{9}+ x^{4}− x + 1 > 0*Proof C*

Consider a polynomial

**x**^{12}− x^{9}+ x^{4}− xIt can be invariantly transformed into

*or*

**x**^{9}·(x^{3}− 1) + x·(x^{3}− 1)*or*

**(x**^{3}− 1)·(x^{9}+ x)

**(x**^{3}− 1)·x·(x^{8}+ 1)This polynomial has only two roots:

*and*

**x=0**

**x=1**As easily checked, values outside interval

*are non-negative and inside this interval the values of a polynomial are negative.*

**(0,1)**Since we are interested in the values of this polynomial

*, the only interval where it's not obvious whether after adding*

**+1***it is positive or not is inside the interval*

**1***.*

**(0,1)**Inside interval

**(0,1)**

**x**

= x^{12}− x^{9}+ x^{4}− x + 1 == x

^{12}+ x^{4}·(1−x^{5}) + (1−x)with every item in parenthesis and every other participant in the above expression is

**positive**, which results in a

**positive**value of an entire expression.

End of proof.

## No comments:

Post a Comment