WORLD OF MATHEMATICS & STATISTICS Some Basic Knowledge Notation and Conventions ℤ = set of integers ℕ = set of natural numbers ℚ = set of rational numbers ℝ = set of real numbers ℂ = set of complex numbers ℝⁿ = Euclidean space of dimension n For a natural number n, the product of all the natural numbers from 1 upto n is denoted by n! For a differentiable function f : ℝ → ℝ, f ′ denotes its derivative. For any natural number n, ℤ/nℤ denotes the ring of integers modulo n. Subsets of ℝⁿ are assumed to carry the induced topology and metric. [a, b] = {x ∈ R : a ≤ x ≤ b} for real numbers a and b with a < b. (a, b) = {x ∈ R : a < x < b} for real numbers a and b with a < b. Some Important Theorems To Remember: Fermat’s Little Theorem: If p is a prime & p is not a divisor of a, then ap-1 ≡1(modp) ⇒ ap ≡ a(modp) ⇒ ak(p-1)+1 ≡ a(modp) , where p is a prime, k,a ∈ ℤ. Wilson’s Theorem: If p be a prime then (p-1)!+1≡ 0(modp). The converse of the theorem is also true. Euler’s Theorem: If n be a positive integer and a is prime to n, then aphi(n) ≡ 1(modn). Dirichlet’s Theorem: There are infinitely many primes of the form 4n+1. Chinese Remainder Theorem: Let m₁,m₂,m₃, . . . . . ,mᵣ denote r positive integers that are relatively prime in pairs, and let a₁,a₂,a₃, . . . . . ,aᵣ denote any r integers. Then the congruences x ≡ ai (mod mi) , i=1(1)r, have common solutions.

Some Tricks & Formulae for Problem Solving : Result : If the sum of two positive quantities is a constant (given),then their product is maximum when two quantities are equal. Ex. Let a+b=12, then max(ab)=36 , i.e. When a equals b. Ex.1. True/False: For x<0 , eⁿ (1 - eⁿ) ≤ 1/4 . Sol:- True. eⁿ,1- eⁿ both are positive. Now eⁿ +(1 - eⁿ)=1 So their product is max when eⁿ =1- eⁿ =1/2,so max[eⁿ(1 - eⁿ)] is 1/4. Ex.2. If a,b are positive real variables whose sum is a constant k, then the minimum value of root of {(1+1/a)(1+1/b)} is A. k-1/k B. k+2/k C. 1+2/k D. none Sol: (C). The given root is minimum when ab is maximum. Now ab is maximum when a=b=k/2. So value of root of{(1+1/a)(1+1/b)}=(1+2/k) Theorem: (de Polinac's formula) Let p be a prime and e be the largest exponent of p such that pe divides n! ,then e=∑ [n/pi], where i = 1(1)∞ . Proof : n!= 1.2. . . . p. . .2p . . . (p2). . .(p+1)p . . .(p3) . . .n ,where [n/p] terms which are divisible by p & among these there are [n/p2] terms which are divisible by p2 and so on. So, the highest power of p that divides n! is [n/p]+[n/p2]+[n/p3]+ . . . +[n/pk] where pk is the largest power of p which is less than or equal to n. Hence we have the proof. Ex.3. How many zeros are at the end of 1000! ? Sol :- The number of 2's is enough to match each 5 to get a 10 . So, [1000/5]+[1000/25]+[1000/125]+[1000/625]=249 . Thus, 1000! ends with 249 zeros. Ex.4. The product of the first 100 positive integers ends with A. 21 zeros B. 22 zeros C. 23 zeros D. 24 zeros Sol:- (D) Put p=5 , n=100,thus from the theorem we have [100/5]+[100/25]=24 zeros as the answer.

Ex.5. The highest power of 3 contained in 100! is A. 46 B. 47 C. 48 D. None Sol:- (C) Highest power of 3 contained in 100! is =[100/3]+[100/3^2]+[100/3^3]+ . . . =33+11+3+1+0+0+. . =48.

INTERMEDIATE VALUE THEOREM:Let [a,b] is a closed and bounded interval and a function f : [a,b] →R be continuous on [a,b]. If f(a) ≠ f(b) then f attains every value between f(a) and f(b) at least once in the open interval (a,b). Proof:- WLG, we assume f(a) < f(b). Let k be a real number such that f(a)R defined by g(x)=f(x)-k, where x is lying in [a,b]. Then g is continuous on [a,b],since f is continuous on [a,b]. g(a) = f(a) − k <0, g(b)=f(b)-k>0. So, g(a) & g(b) are of opposite signs,by Bolzano's theorem there are at least one point c in (a,b) such that g(c)=0. Therefore, f(c)=k. The proof is complete. Ex.6. A function f :[0,1]->[0,1] is continuous on [0,1]. Prove that there exists a point c in [0,1],such that f(c)=c. Sol:- If f(0)=0 or f(1)=1 the existance is proved. We assume f(0)≠0 & f(1)≠1. Let us consider a function g :[0,1]->R defined by g(x)=f(x)-x , x in [0,1] . g is continuous on [0,1] & g(0)=f(0)>0 , since f(0) belongs to [0,1] & f(0)≠0. Also g(1)=f(1)-1<0,since f(1) belongs to [0,1] & f(1)≠1 . By IVT there exists a point c in (0,1) such that g(c)=0. Therefore, f(c) = c.

Result:- If f is a continuous function that satisfies f(x+y)=f(x)+f(y) for all x,y; then f(x) = xf(1). Proof:- (Hints) We know that if f is a function satisfying f(x+y)=f(x)+f(y) for all x,y then f(x)=kx ,where k is a constant. [To show this, Check the differentiability first. Then f '(x)= Lt {f(x+h)-f}/h ,where h->0. ⇒ f '(x)= Lt {f(h)/h} where h->0 = f '(0)=k(say). Then integrate both sides, you will get f(x)=kx+c ------(1) Put x=y=0 in the given equation,then f(0)=0. From (1), c=0, i.e. f(x)=kx. ] Here f(1)=k ,so f(x) = kx = xf(1). Ex.7. Let g be a continuous function with g(1)=1 such that g(x+y) = 5g(x) g(y) for all x,y. Find g(x) Sol:- g(x+y) = 5g(x)g(y) 5g(x+y) = 5g(x)5g(y), taking logarithm in bothsides ⇒ log[5g(x+y)] = log[5g(x)] + log[5g(y)] ⇒ f(x+y)=f(x)+f(y), taking f(x+y) = log[5g(x+y)] ⇒ f(x)= xf(1) [By the above result] So, log[5g(x)] = x.log[5g(1)] ⇒ 5g(x) = 5x ,since g(1)=1 (given) ⇒ g(x) = 5x-1 . Result: Considering Cauchy's functional equation:- f(x+y) = f(x) + f(y) ; Generally while you are solving a problem of functional equation with two variables x & y then in most of the situations to solve a given problem you are to put these values Put x= -x Put y= -x Put x= f(y) Put x= y Put x= f(x) etc . . . The choices differ & it completely depends on the given problem which requires these selections.

Ex.8. If f(x+y) = f(x) + f(y) for all y in ℝ. Then f(x) is a function which is A. Odd B. Even C. None Sol:-(A) Putting x= -x ,then f(x)+f(-x)=f(0) -----(1) Again put x=0 in (1), so f(0)=0. So, from (1), f(-x)= -f(x) which is an odd function. Ex.9. If the function f satisfies the relation f(x+y)+f(x-y) = 2f(x).f(y) for all x,y in ℝ. And f(0)≠0. Prove that f(x) is even function. Sol:- f(x+y) + f(x-y) = 2f(x).f(y) ---(1) Replace x by y & y by x Then f(y+x) + f(y-x) = 2f(y)f(x) ---(2) Solving (1) & (2), we get f(y-x)= f(x-y) Put y=2x then f(x) = f(-x). Hence f(x) is an even function. Ex.10. If f(x) is a real valued fiction such that 2f(x) + 3f(-x) = 15 - 4x , for all x. Then f(2) is A. 15 B. 22 C. 11 D. 0 Sol:- (C) Given 2f(x) + 3f(-x) = 15 - 4x -----(1) put x = -x in the given functional equation (1), we have 2f(-x) + 3f(x) = 15 + 4x ------------(2) Solving (1) & (2) we have f(x) = 3 + 4x . So, f(2)=11. EULER'S PHI FUNCTION:The number of integers ≤ n and co-prime to n is called Euler's function for n & is denoted by ϕ(n). Ex. I. ϕ(1)=1 (since 1 is the only integer ≤1 and co-prime to 1) II. ϕ(8)=4 (since 1,3,5,7 are the only four integers <8 and co-prime to 8) III. ϕ(12)=4 (since co-prime integers less than 12 are 1,5,7,11). Properties:I. ϕ(n) is a multiplicative function. II. For n>2, ϕ(n) is an even integer. (Proof:- If (a,n)=1 then (n-a,n)=1. Thus integers co-prime to n occur in pairs of type a and n-a. So, ϕ(n) is even.)

Note:I. If p is prime, then ϕ(p) = p-1. e.g. ϕ(2)=1, ϕ(3)=2, ϕ(7)=6. II. If a,b,c,d, . . . are positive integers prime to each other, then ϕ(abcd . . .)=ϕ(a)ϕ(b)ϕ(c)ϕ(d) . . . . . . e.g. ϕ(30)= ϕ(2) ϕ(3) ϕ(5)=1.2.4=8. III. If n = aq.br.cs…….pz is the prime factorization, where a,b,c,…,p are primes, then ϕ(n) = n(1- 1/a)(1- 1/b)(1- 1/c). . . . .(1- 1/p). e.g. ϕ(360)=360(1-1/2)(1-1/3)(1-1/5)=96. IV. If p is a prime and n>0 then ϕ(pⁿ) = pⁿ (1 – 1/p)= pⁿ - pⁿ-1 . e.g. ϕ(9)= ϕ(32)= 32 – 3=6. V. The sum of positive integers less than n & relatively prime to n is 1/2. n ϕ(n), where n>1. Ex.11. Calculate ϕ(1001). Sol:- 1001=13*7*11,where the numbers are relatively prime to each other. So, ϕ(1001)=ϕ(13)ϕ(7)ϕ(11) =12.6.10=720. (since 13,7,11 are all prime numbers) Application of Euler's function in abstract algebra:It uses to determine the number of generators of a finite cyclic group. Ques: How many generators of the cyclic group ℤn are there? Is this same as the number of generators of any othere cyclic group of order n ? Ans: A number in ℤn generates ℤn if and only if it is relatively prime to n . Hence, since there are ϕ(n) integers in ℤn that are relatively prime to n, then there are ϕ(n) generators for ℤn. Finding No of Generators of a Cyclic Group: The total number of generators of a finite cyclic group of order n is ϕ(n). Finding No of Subgroups of a Cyclic Group: If n=aq.br.cs…….pz is the prime factorization the Total No Of Subgroups=τ(n)=(q+1)(r+1)(s+1)……(z+1).

Examples:I. The number of generators of the cyclic group (S, . ) where S={1,i,-1,-i} is 2, since ϕ(4)=2. II. The number of generators of the cyclic group of a prime order p is p-1 since ϕ(p)=p-1 Ex.12. A cyclic group of order 60 has A.12 B.15 C.16 D.20 generators. Sol:- (C) We need to find ϕ(60) ϕ(60) = ϕ(3)ϕ(4)ϕ(5) = 2.2.4 = 16, Since 3,4,5 are relative primes. And ϕ(4)=4(1 -1/2)=2 Since 3,5 are primes so ϕ(3)=2,ϕ(5)=4. Ex.13. How many subgroups are there of a cyclic group of order 100 ? Sol:- 100=22.52 where 2,5 are primes. So, Total No Of Subgroups=τ(n)=(2+1)2=9. Theorem: Every cyclic group is Abelian. Proof: Suppose that G is a cyclic group that is generated by the element g. Let x and y be arbitrary elements of G. we must show that xy = yx. Since G is generated by g, there must exist integers r and s such that x = gr, y = gs. But then xy = gr.gs = gr+s= gs. gr = yx. But the Converse of the theorem is not necessarily true. An Abelian Group is said to be Cyclic if the order of the abelian group is prime or product of primes or power of primes. Important Notes: The order of cyclic group is same as the order of its generators. Every group of Composite Order is not Cyclic. Every group of order p2 is abelian, where p is a prine.

Ex.14. For each n, 9 ≤ n ≤ 16, answer the following questions: (a): Is every group of order n cyclic? (b): Is every group of order n abelian? (c): Is every abelian group of order n cyclic? Answer: Consider n = 9. Then, since 3 is not relatively prime to itself, ℤ/3 × ℤ/3 is not cyclic, so we see that not every group of order 9 is cyclic and not every abelian group of order 9 is cyclic. Since these are the only abelian groups of order 9 and we know that groups of order p2 for a prime p are abelian, this comprises the entire set of groups of order 9, so we can say that every group of order 9 is abelian. Consider n = 10. Then ℤ/10 ≅ ℤ/2 × ℤ/5 since (2, 5) = 1 and, since these are the only abelian groups, we see that every abelian group of order 10 is cyclic. Now, D₅= < x, y| x⁵ = y2 = 1, yxy⁻¹ = x⁻¹ > is of order 10 but isn’t abelian, so we see that every group of order 10 is not abelian and, thus, not all groups of order 10 are cyclic. Consider n = 11. Then every subgroup of a group of order 11 must be of order 1 or 11, so each element except the identity is of order 11, so the only group of order 11 is ℤ/11. Consider n = 12. Then, since (3, 4) = 1, ℤ/12 ≅ ℤ/3 × ℤ/4. However, since ℤ/4 is not isomorphic to ℤ/2×ℤ/2, we see that ℤ/12 is not isomorphic to ℤ/3×ℤ/2×ℤ/2, which isn’t cyclic, so we know that not every group of order 12 is cyclic and not every abelian group of order 12 is cyclic. Furthermore, D₆ = < x, y| x⁶ = y2 = 1, yxy⁻¹ = x⁻¹ > is of order 12 and non-abelian, so we see that not every group of order 12 is abelian. Consider n = 13. By the same reasoning given in the case where n = 11, we see that ℤ/13 is the only group of order 13, so the answer to all the questions is “yes”.

Consider n = 14. Since (2, 7) = 1, ℤ/14 ≅ ℤ/2 × ℤ/7 and, since these are the only abelian groups of order 14, we see that every abelian group of order 14 is cyclic. Now, D₇ = < x, y| x⁷ = y2 = 1, yxy⁻¹ = x⁻¹ > is a non-abelian group of order 14, so not every group of order 14 is abelian or cyclic. Consider n = 15. Since (3, 5) = 1, ℤ/15 ≅ ℤ/3 × ℤ/5 and, since these are the only abelian groups of order 15, we see that every abelian group of order 15 is cyclic. Furthermore, since 3 does not divide (5 − 1) = 4, we can conclude that any group of order 15 is abelian and, therefore, it is cyclic. Consider n = 16. Then ℤ/16 ≠ ℤ/4 × ℤ/4, so not every abelian group of order 16 is cyclic. Furthermore, D₈ = < x, y| x⁸ = y2 = 1, yxy⁻¹ = x⁻¹ > is a non-abelian group of order 16, we see that not every group of order 16 is abelian. Pigeon Hole Principle (P.H.P.)/Dirichlet's Box Principle: Introduction: If 4 pigeons fly into 3 pigeon-holes, then some pigeon hole(s) must receive at least 2 pigeons. The statement is an obvious one. Theorem:- If n+1 objects are distributed into n boxes, then at least one box has at least two objects. A more generalised form of this principle is as follows : If N pigeons are to be placed into K boxes, then there is at least one box containing at least [N/K] objects, where [x] stands for greatest integer function of x. The proof is an obvious one. Let us illustrate these through examples.

Ex.15. How many students, each of whom comes from the one of 50 states, must be enrolled in a college to guarantee that there are at least 100 who comes from the same state? Ans: Here we have to find the number of students (or, pigeons) say N, so that at least 100 come from the same state (or, pigeon holes) ,i.e., number of pigeon holes is 50, then by the generalized form of P.H.P. [N/50] = 100 => N/50 > 99 => N > 99×50 = 4950 Therefore, N=4951 is the required number of students. Ex.16. A box contains three pair of socks; coloured red, blue & white. Suppose I take out the socks without looking at them. How many socks must I take out in order to be sure that they will include a matching pair? Ans: If I take only 2 or 3 socks, it is possible that they are all different. For example, they may be one red one blue or one red, one blue & one white. But if I take 4 socks, there must include a matching pair. Here the 4 chosen socks are the "objects" & 3 colours are the "boxes". So by P.H.P. at least two of the 4 chosen socks must have the same colour & hence must for a matching pair. Thus the minimum number of socks to be drawn is 4. Ex.17. Show that in a group of 8 people, at least two will have their birthday on the same day of the week. Ans: Here the 8 peoples are the "objects" & 7 days are the "boxes". Hence by P.H.P. at least two of the 8 people must belong to the same day. Similarly, it is clear that in a group of 13 people, at least two will have their birthday in the same month. Ex.18. Two boxes contain between them 65 balls of several different sizes. Each ball is white,black,red or yellow. If you take any 5 balls of the same colour, at least two of them will always be of the same size. Prove that there are at least 3 balls which lie in the same box, have the same colour & are of the same size. Ans: Here we will make repeated use of P.H.P. As there are 65 balls & 2 boxes, so one of these boxes will contain at least [65/2] + 1 = 33 balls. Consider that box, now we have four colours, so there must be at least

[33/4]+1 = 9 balls of the same colour. There can be at most four different sizes available for these 9 balls of the same colour ( see 3 rd line of the question). Thus for these 9 balls (of the same colour & in the same box) there must be at lest [9/4]+1=3 balls of the same size (as at most 4 different sizes are available). Ex.19. Prove that no 7 integers, not exceeding 24, can have the sum of all subsets different. Ans : Let S be any 7-subset of {1,2, . . . ,24}. The number of non-empty subsets of S having at most 4 elements is 7C1 + 7C2 + 7C3 + 7C4 = 98. If T is any one of these subsets,then the sum of elements in T is between 1 and 21+22+23+24=90. Since 90 < 98, by P.H.P. it follows that the sums corresponding to the above 98 subsets can't all be different. Remark : While using P.H.P. the main step is to decide what are the "objects" and what are the "boxes" . A clue is provided by the fact that the arithmetic of the numbers involved in the problem is to be exploited. DESCARTE'S RULE OF SIGNS (1) The maximum number of positive real roots of a polynomial equation f(x)=0 is the number of changes of signs from positive to negative & negative to positive in f(x). (2) The number of negative roots of f(x)=0 is the number of changes of signs in f(-x)=0. USEFUL THEOREM:- Let f(x) be a real polynomial of degree n(≥1) and a,b be two real numbers such that a
Remark:- Every odd degree polynomial has at least one real root since complex roots occur in pairs. Ex.20. If a>0, prove that x3+ax+b =0 has two complex roots. Sol:- Case-1: b>0 Let f(x)= x3+ax+b Signs are +,+,+ .Since there is no sign change in f(x). Thus, f(x)=0 has no positive root. Again, f(-x)= -x3-ax+b Signs are -,-,+ Since there is one change of sign in f(-x) So, f(x)=0 has one -ve real root & two complex roots. Case-2:- b<0 Check yourself. Case-3:- b=0 so,we have ax +x3 =0 So, x=0, x2+a=0 so,it has two complex roots. Theorem: The number of ways to divide n identical things among r persons when each get at least one is (n-1)C(r-1) . Example: 10 identical balls can be put in 4 different boxes in a row such that no box remains empty, is (10-1)C(4-1) =9C3 . RESULT: The number of solutions of the equation x+y+z+ . . . . +k = n for positive integers (each x,y,z, . . .,k > 0) with total number of element=r, is (n1)C(r-1) . Theorem: The total number of ways of dividing n identical things among r persons, each one of whom can receive 0,1,2,3 or more things (≤ n) is (n+r1)C(r-1) . Example: 30 mangoes can be distributed in 5 boys in (30+5-1)C(5-1) ways. RESULT: The number of solutions of the equation x+y+z+ . . . . +k = n for nonnegative integers ( each x,y,z, . . .,k ≥ 0 ) with total number of element = r, is (n+r-1)C(r-1) .

Ex.21. How many non-negative integer solutions are there of the equation x+y+z=18. Sol:- Use the Result-2 given above. Here n=18, r=3 . So the answer is = 20C2. Result: If k is an eigenvalue of A, then kⁿ is an eigen value of Aⁿ , for any positive integer n. Proof:- Since k is an eigenvalue of A. For x ≠ 0, Ax=kx => A(Ax)=A(kx) => A²x = k²x => k² is an eigenvalue of A². We have, A²x=k²x => A(A²x)=A(k²x) => A³x = k³ => k³ is an eigenvalue of A³. In the same way, we can prove that for any n = 2,3,4, . . . ; kⁿ is an eigenvalue of Aⁿ. Observation: If a square matrix has 0 as its eigenvalue then the matrix is singular. Justification:- | A-k.I |=0 is the characteristic equation, where k is the eigenvalue of A. If k=0, then |A|=0. Ex.22. Let A be a 3×3 matrix with eigenvalues 1,−1 and 3. Then A. A² + A is non-singular B. A² − A is non-singular C. A² + 3A is non-singular D. A² − 3A is non-singular Sol:- (C) Eigenvalues of A is 1,−1,3. So, eigenvalues of A² is 1,1,9 => eigenvalues of A² + 3A is 1+3.1, 1+3(−1), 9+3(3), i.e., 4, −2 and 18. As no eigenvalue is 0 ,so A² + 3A is non-singular. Result: If A be an n×n non-singular matrix, then |adjA|=|A|ⁿ-1. Proof: As we know that A.adjA = |A|.I or, det{ A.adjA }=det{ |A|.I }

or, |A|.|adjA|= |A|ⁿ Using det(AB)=det(A).det(B) and det(|A|.I)=|A|ⁿ.I Hence we get the result. [ As |A|≠0 ]. Ex.23. Let A be a 3×3 real matrix with det(A)=6. Then find det(adjA). Sol:- Using the above result, putting n=3 So, det(adjA) = 6² =36. Cycle in the Last Digit The last digit (the ones place) of a decimal integer d is the remainder of the division d/10. Equivalently, the last digit is the result of the operation d mod 10, when following the convention that the least nonnegative value the common residue is returned. Modular arithmetic, combined with iterative generation of the positive powers of two, allows us to show the cycle in the last digit: Let us check the cycle of 2n, n ≥ 1, cycle through the four ending digits 2, 4, 8, and 6. 1. 2¹ ≡ 2 (mod 10) 2. 22 ≡ 4 (mod 10) 3. 23 ≡ 8 (mod 10) 4. 2⁴ ≡ 6 (mod 10) 5. 2⁵ ≡ 2 (mod 10) 6. . . . . . . . The cycle implies that powers of two with the same ending digit are related, their exponents differing by a multiple of four:  Ends in 2: 21, 25, 29, 213, 217, … .  Ends in 4: 22, 26, 210, 214, 218, … .  Ends in 8: 23, 27, 211, 215, 219, … .  Ends in 6: 24, 28, 212, 216, 220, … . You can express these relationships more succinctly using the laws of exponents, showing explicitly that the ending digit of the first four positive powers of two determine the ending digit of all positive powers of two:  Ends in 2: 21·24k, or 21+4k, k ≥ 0.  Ends in 4: 22·24k, or 22+4k, k ≥ 0.

Ends in 8: 23·24k, or 23+4k, k ≥ 0.  Ends in 6: 24·24k, or 24+4k, k ≥ 0. More explicitly we can write : Ends in 4: 210, 230, 250, 270, 290, . . . . . . Ends in 6: 220, 240, 260, 280, 2100, . . . . . 

Ex.24. The last digit of 280 is (a) 2 (b) 4 (c) 6 (d) 8 Solution: (c) Apply this trick here: Ends in 6: 24·24k, or 24+4k, k ≥ 0. Put k=19. Theorem: In an equation with real coefficients, non-real complex roots occur in conjugate pairs. Corollary: Every equation of odd degree having real coefficients, has at least one real root, because complex roots occur in pairs. Theorem: In an equation with rational coefficients irrational roots occur in conjugate pairs. Theorem: α is a repeated roots of f(x)=0 iff α is a common root of f(x)=0 & f′(x)=0. A complex number α is a repeated root of the polynomial equation f(x)=0 iff α is a root of d(x)=0, where d(x)=(f(x), f′(x)). Ex.25. If the roots of the equation xⁿ-1=0 are 1, α, β, γ,……., then show that (1- α)(1- β)(1- γ)……..= n. Solution: xⁿ-1= (x-1)(x- α)(x- β)(x- γ)……., dividing both sides by (x-1). ⇒ xn-1+ xn-2+. . . . . . . . +x2 + x+1=(x- α)(x- β)(x- γ)……. Putting x=1, we get the required answer. Relationship between roots & coefficients:1. If α, β be the roots of the equation ax2+bx+c=0 then α+β= -b/a & αβ=c/a. 2. If α, β, γ be the roots of the equation ax3+bx2+cx+d=0 then α+β+γ=-b/a, αβ+αγ+βγ= c/a, αβγ=d/a.