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CauchySchwarz states that:
(x1² + x2² + x3²)(y1² + y2² + y3²) ≥ (x1y1 + x2y2 + x3y3)², for real xi and yjProblem 1:
If a, b, c ,d are real numbers with a² + b² + c² + d² = 100, then what is the maximum value of 2a + 3b + 6c + 24d.
Ans > Using CauchySchwarz, we can say
(a² + b² + c² + d²)(2² + 3² + 6² + 24²) ≥ (2a + 3b + 6c + 24d)²
(100)(625) ≥ (2a + 3b + 6c + 24d)²
So, 2a + 3b + 6c + 24d ≤ 250Problem 2:
Find the least value of x² + 4y² + 9z², if x + y + z = 14
Ans > {x² + (2y)² + (3z)²}{1 + (1/2)² + (1/3)²} ≥ {x + 2y(1/2) + 3z(1/3)}²
=> (x² + 4y² + 9z²) ≥ (x + y + z)²/{1 + (1/2) + (1/3)}²
=> (x² + 4y² + 9z²) ≥ 196/(49/36)
=> (x² + 4y² + 9z²) ≥ 144Problem 3:
If x² + y²  6x + 4y = 4, find the maximum value of 3x + 4y.
Ans > Given equation is: (x3)² + (y+2)² = 9
Using Cauchy Schwarz, we get
[(x  3)² + (y + 2)²][3² + 4²] ≥ [3(x  3) + 4(y + 2]²
9 * 25 ≥ (3x + 4y 1)²
3x + 4y 1 ≤ 15
3x + 4y ≤ 16

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
A brief video detailing the Algebra AMGM Concept in a Question

If a + b + c + d = 20, how many unique, nonnegative integer solutions exist for (a, b, c, d)?
Let us try and understand the concept behind solving such questions.
This is like distributing 20 identical chocolates between 4 kids A, B, C, and D. You arrange these 20 chocolates in a line
C C C C … C
Now, you put in partitions in between them. Let me denote the partitions with P
C C C P C P C C C C C P C C C C C C C C C C C
A gets the chocolates before the first partition, B gets the chocolates between the first two partitions, C gets the chocolates between the second and the third partition and D gets the chocolate after the third partition. In the arrangement shown above A gets 3 chocolates, B gets 1, C gets 5, and D gets 11. Any rearrangement of the above, will lead to a new distribution of chocolates.
The above can be rearranged in 23! / 20! 3! We got this because there are a total of 23 entities out of which 20 Cs are identical and 3 Ps are identical.
Now, to extend this concept, what if we have to distribute ‘n’ chocolates in ‘r’ kids. After putting the n chocolates in a line, we would need r  1 partitions. This would mean that there will be a total of n+r1 entities out of which n would be identical (of one type) and the other r  1 would be identical as well (of another type). So, the number of ways in which that would be possible = (n + r  1)! / n! (r1)!
This, in other words, is (n + r  1) C (r  1)
And now, we can use this formula to solve any similar questions
a + b + c + d = 20
Case 1: a, b, c, d are nonnegative integers.
Number of solutions = (20 + 4  1) C (4  1) = 23 C 3 = 1771
Case 2: a, b, c, d are positive integers
We allocate at least a value of 1 to a, b, c, d.
So, we can say a = a’ + 1, b = b’ + 1, c = c’ + 1, d = d’ + 1 where a’, b’ c’, d’ are nonnegative integers
=> a’ + 1 + b’ + 1 + c’ + 1 + d’ + 1 = 20
=> a’ + b’ + c’ + d’ = 16
=> Number of solutions = (16 + 4  1) C (4  1) = 19 C 3
Case 3: a, b, c, d are nonnegative integers such that a > 5 and b > 2
We allocate at least a value of 5 to a and 2 to b
So, a = a’ + 5 and b = b’ + 2
=> a’ + 5 + b’ + 2 + c + d = 20
=> a’ + b’ + c + d = 13
=> Number of solutions = (13 + 4  1) C ( 4  1) = 16 C 3
Case 4: a, b, c, d are nonnegative integer such that a > b
Let us first consider the situation where a = b
If a = b = 0, c + d = 20. This has 21 solutions
If a = b = 1, c + d = 18. This has 19 solutions
If a = b = 2, c + d = 16. This has 17 solutions
.
.
If a = b = 10, c + d = 0. This has 1 solution
So, the total number of a solutions when a = b is 21 + 19 + 17 … + 1 = 11/2*(21 + 1) = 121
We know that the number of solutions when a, b, c, and d are nonnegative integers is 1771. Out of these 1771 cases, in 121 cases a = b.
So, in 1771  121 = 1650 cases a is not equal to b.
In half of the above cases a will be greater than b whereas in the other half of the cases a will be less than b.
So, number of solutions where a > b is 1650/2 = 825

Equation type: Ax + By = C
Few rules to find integral solutions of this type of equations.
First, reduce the equation in lowest reducible form.
After reducing, if coefficients of x and y still have a common factor, the equation will have no solutions.
If x and y are coprime in the lowest reducible form, find any one integral solution. The rest of the solutions can be derived from that integral solution.
For each successive integral solutions of the equation, the value x and y will change by a coefficient of the other variable .If the equation is of the type Ax – By=C (after getting the lowest reducible form) ,an increase in x will cause increase in y .If the equation is of the type Ax + By=C,an increase in x will cause a decrease in y.Let us take an example.
2x + 3y = 39.
Step1.The equation is already in its reduced form and we can see that coefficients of x and y are coprime.
Step2.For a given equation, you should start substituting values (by hit and trial) for the variable that has larger coefficient to find out first integral solution. In this case, it is y. Now, if we take y = 0, we will get x = 39/2(not an integer). Again, if we take y=1, we will get x = 18. So, (18,1) is our first solution.
Step3.If you understand the 4th point mentioned above, at one of any two consecutive integral values of y, the value of x will come out to be an integer OR at one of the 3 consecutive values of x, the value of ywill come out to be an integer. That means, if we add 2n (where n is an integer) to the first value for y, we will have to subtract 3n from the first value of x to get integral solutions. That means,
If y =1 +2(1) = 3 , x= 183(1) = 15.
If y= 1 + 2(2) = 5, x= 18 – 3(2) = 12.
If y= 1 + 2(3) = 7, x = 18 – 3(3) = 9 and so on.Step4.This equation will have infinite number of integral solutions but finite number of nonnegative integral solutions. Let’s see how we can find it.
We can keep increasing the value of y in the positive direction but x will be decreasing simultaneously and become less than 0 at one point. As lowest non negative integral value of y is 1,highest allowable positive value of x is 18 and it is decreasing by 3. So, x can take 7 non negative integral values and they are 18, 15, 12, 9, 6, 3 and 0.Hence the given equation has 7 non negative integral values.
Note: In equation Ax + By = C, if C is divisible by any of A or B, then number of nonnegative integral solutions = {C/LCM(A,B)} + 1

Equation type: x + y = n
Let x = p and y = q, then positive integral solutions= n1C21= n1.
Now, for each solution (x1,y1), there would exist 4 values for x and y, They are >
(x1,y1), (x1,y1), (x1,y1) and (x1,y1).
Therefore, total number of positive integral solutions = 4(n1).

Type 1 : x + y = n
Total solutions = 4n.Example : x + y = 5.
Total solutions = 4 x 5 = 20.Type 2 : x + y + z = n.
Total solutions = 4n^2 + 2.Example : x + y + z = 15.
Total solutions = 4 x 15^2 + 2 = 902.Type 3 : x + y + z + w = n.
Total solutions = (8/3)n(n^2 +2).Example : x + y + z + w = 9.
Total solutions = (8/3) x 3 x (3^2 +2 )
= 8 x 11
= 88.Type 4 : ax + by = n.
Non negative solutions = n/LCM (a,b) + 1. if either a or b is divided by n.
For positive solutions just remove x=0,y=0 from non negative solution.Example : 2x + 3y = 30.
positive integral solutions
= 30 / LCM (2,3)  1
= 5  1 = 4.Read more @ https://www.mbatious.com/topic/181/numberofintegralnonnegativepositivesolutionsvikassaini

Type : a x b = N
no of positive solution = no of factors of N.
no of integrated solution =2 x no of factors of N.Example : a x b = 36.
36 = 2^2 x 3^2.
total no of factors = (2+1)(2+1) = 9.
positive solution = 9.
total solution = 2 x 9 = 18.
Here multiply by 2 because even negative sign also contains in total solution.

Fundamental Theorem of Algebra : A polynomial of degree n can have at most n distinct real roots.
Descartes' Rule of Signs : It tells us that the number of positive real roots in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. The number of negative real roots of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(x) or less than this by an even number.
Example : Determine the number of positive and negative real roots for the given function f(x) = x^5 + 4x^4 − 3x^2 + x − 6
Our function is arranged in descending powers of the variable, if it were not we would have to do that as a first step. Second we count the number of changes in sign for the coefficients of f(x).
Here are the coefficients of our variable in f(x): 1 + 4 − 3 + 1 − 6
Our variables goes from positive(1) to positive(4) to negative(3) to positive(1) to negative(6).
Between the first two coefficients there are no change in signs but between our second and third we have our first change, then between our third and fourth we have our second change and between our 4th and 5th coefficients we have a third change of coefficients.Descartes´ rule of signs tells us that the we then have exactly 3 real positive roots or less than by it an even number. Hence our number of positive roots must then be either 3 or 1 ( = 3  2).
In order to find the number of negative zeros we find f(x) and count the number of changes in sign for the coefficients:
f(−x) = (−x)^5 + 4(−x)^4−3(−x)^2 + (−x) − 6 = −x^5 + 4x^4−3x^2 − x − 6
Here we can see that we have two changes of signs, hence we have two negative roots or less than it by an even number.
Totally we have 3 or 1 positive roots or 2 or 0 negative roots.
Descartes' Rule of Signs does not provide the exact number of positive and negative real roots.
Refer : https://www.mathplanet.com/education/algebra2/polynomialfunctions/descartesruleofsign