UNIT-1 SQUARE ROOT EXERCISE 1.1.1

Transcript

1 UNIT-1 SQUARE ROOT EXERCISE Find the square root of the following numbers by the factorization method (i) x 3 4 = (2 5 ) 2 x (3 2 ) = = (2 5 ) 2 (3 2 ) 2 = 2 5 x 3 2 = 288

2 (ii) = 22 X = = (2 197) 2 = 2 X 197 = 394 (iii) = = = 3 x 47 = 141

3 2. Find the square root of the following numbers. (i) x = = (23 ) = (2 3 ) = 8x17 10 = = 13.6

4 (ii) = = = ( ) = = 4.42

5 3. Exploration: Find the squares of the numbers 9, 10, 99, 100, 1000, and 9999.Tabulate these numbers. How many digits are there in the squares of a numbers when it has even number of digits and odd number of digits? 9 2 = 81 digits = 100 digits = 9801 digits = digits = digits = digits = digits-8 They follow the rule 2n if the number has even digits 2n-1 if the number has odd digits 4. If a perfect square A has A digits, how many digits to you expect in A If a perfect square A has n digits then, A has n 2 Digits if n is even (n+1) 2 Digits if n is odd.

6 EXERCISE Find the square root of the following number by division method: (i) = 73 (ii) = 137

7 (iii) = 168 (iv) = 432

8 2. Find the least number to be added to get a perfect square. (i) > 78 2 Therefore we find = is the least number to be added to 41 get a perfect square. (ii)

9 113 2 = is the least number to be added to to make it a perfect square. (iii) = is the least number to be added to to make it a perfect square.

10 (iv) lies between and = = is to be added to make it a perfect square. 3. Find the least number to be subtracted from the following number to get a perfect square: (i) is the least number to be the least number to be subtracted to make 1234 a perfect square.

11 (ii) should be the least number to be subtracted to make 4321 a perfect square. (iii) should be the least number to be subtracted to make a perfect square.

12 (iv) should be the least number to be subtracted to make a perfect square. 4. Find the two consecutive perfect square numbers between which the following number occurs. (i) The number 4567 lies between 67 2 and = = 4624

13 (ii) lies between the squares of and (iii) = = lies between the squares of and = = 89401

14 (iv) lies between and = = A person has 3 rectangular plots of dimension 112m x 54m, 84m x 68m and 140m x 87m. In different places. He wants to sell all of them and buy a square plot of maximum possible area approximately equal to them to the sum of these plots. What would be the dimensions of such a square plot? How much lands he has to be lossed? The area of the 3 rectangular plots is (112 x 54) m = 6048 sq. m. (84 x 68) m = 5712 sq. m. (140 x 87) m = sq. m. Total area = sq. m.

15 The area of the new square plot is = sq. m. The person would have to loose area of sq. m.

16 EXERCISE How many digits are there after the decimal points in the following? (i) (3.16) 2 (3.16) 2 = Ans: 4 (ii) (1.234) 2 (1.234) 2 = Ans: 6 (iii) (0.0023) 2 (0.0023) 2 = Ans: 8 (iv) (1.001) 2 (1.001) 2 = Ans: 6 2. Given that the numbers below are all squares of some decimal numbers, how many digits do you expect in their square roots? (i) = 9.21 Ans: 3

17 (ii) (iii) = Ans: = Ans: 4 (iv) = Ans: 5 3. Find the square root of the following number using division method. (i) = 25.53

18 (ii) = (iii) = 9.77

19 (iv) = A square garden has an are m 2. A trench of one meter wide has to be dug along the boundary inside the garden. After digging the trench, what will be the area of the out garden? Area of the square garden is m 2 Length of a side of the garden the is in m.

20 If a trench of one meter is dug along the boundary inside the garden the length of a side of the garden will be ( ) m. = m. 1z zxdrjhzkgfjh 1m The area of the remaining garden is (155.12) = m 2

21 EXERCISE Round off the following numbers to 3 decimal places Find the square root of the following numbers correct to 3 decimal places. (i) =

22 (ii) = = (iii) =

23 (iv) = (v) =

24 (vi) = Find the square root of the following numbers correct to 4 decimal places. (i) =

25 (ii) =

26 (iii) =

27 4. Find the approximation from below to 4 decimal places to the square root of the following numbers. (i) = ( ) 2 = <5< = ( ) is the approximation from below of 5 to 5 decimal places.

28 (ii) = ( ) 2 = < 8 < = ( ) is the approximation from below of 8, correct to 5 decimal places.

29 5. A square garden has area of 900M 2. Additional land measuring equal area, surrounding it, has been added to it. If the resulting plot is also in the form of a square, what is its side correct to 3 decimal places? Ares of square garden is 900m 2. If additional land, measuring equal area to added to the garden, its area will be ( ) 2 = 1800m 2 Side of the new garden is 1800 m. A = side x side = (side) 2 = 42.43m

30 NON-TEXTUAL QUESTIONS 1. Find the square root of the following numbers. (i) = = (2x221 ) = = 4.42 (ii) = (4 353 ) = (100 0 )2 = =

31 2. If a perfect square A has n digits, how many digits do you expect in A? If a perfect square A has n digits then, A has n 2 Digits if n is even (n + 1) 2 Digits if n are odd. 3. Find the square root (i) = 615 (ii) = 961

32 4. Find the least number to be added to get the perfect square. (i) > Therefore we find = should be added to to make it a perfect square. (ii) x x x

33 = should be added to to make it a perfect square.

34 ADDITIONAL QUESTIONS 1. There are 500 students in a school. For the sports day display they have to arrange themselves so as to have number of rows equal to number of columns. How many children would be left out in this arrangement? Ans: since number of rows should be equal to number of column rows xxx 500 x 2 = row and column are possible 22 x 22 = 484 students can perform the drill. Remaining students = 16students 2. Students of class IX wanted to rise the bund to help the needy student of their school. Each student donated as many rupees as the number of students in the class and the amount collected was 6044 find the number of students. Ans: Let the number of students = x Then each students will donates = rupees x Amount collected xxx = x 2 x 2 = 6044 x = 6044 = 78 No. of students in that class = 78

35 3. Find the smallest square number that is divisible by each of the number, 8, 15 and 20. Ans: The smallest number divisible by 8, 15, and 20 is their LCM. L.C.M of 8, 15, and 20 is 120. But 120 is not a square number 120 = 2 x 2 x 2 x 3 x So to make it perfect square we have to multiply by 2 x 3 x 5 Then = 2x2 x 2x3x5 x (2x3x5) = 2x2 x 2x2 x 3x3 x 5x5 = = is a perfect number that is divisible by 8, 15, and 20.

36 4. Express 13 as sum of two consecutive integer we have n = 13. Ans: n = = = = 84 n = = = = = = can be expressed as sum of 84 and 85

37 UNIT-1 MULTIPLICTION OF POLYNOMIALS EXERCISE Evaluate the following products: (i) ax 2 ( bx + c) = ax 2 (bx) + ax 2 (c) = abx 3 + acx 2 (ii) ab (a+b) = ab (a) + ab (b) = a 2 b + ab 2 (iii) a 2 b 2 (ab 2 +a 2 b) = a 2 b 2 (ab 2 ) + a 2 b 2 (a 2 b) = a 3 b 4 + a 4 b 3 (iv) b 4 (b 6 + b 8 ) = b 4 (b 6 ) + b 4 (b 8 ) = b 10 + b 12

38 2. Evaluate the following products: (i) (x+3) (x+2) = (x+3) x + (x+2) x = x 2 + 3x + 2x + 6 = x 2 + 5x + 6 (ii) (x+5) (x 2) = (x+5) x + (x+5) ( 2) = x 2 +5x 2x 10 = x 2 + 3x 10 (iii) ( y 4 ) ( y + 6 ) = (y 4) y +(y 4)6 = y 2 4y + 6y 24 = y 2 + 2y 24 (iv) (a 5) (a 6) = (a 5) a + (a 6) ( 6) = a 2 5a 6a + 30 = a 2 11a +30

39 (v) (2x+1) (2x 3) = (2x+1)2x + (2x+1) ( 3) = 4x 2 + 2x 6x 3x = 4x 2 4x 3 (vi) ( a + b ) ( c + d ) = (a + b) c + (a+b) d = ac + bc + ad + bd (vii) ( 2x 3y ) ( x y ) = (2x 3y) x + (2x 3y) ( y) = 2x 2 3xy 2xy +3y 2 = 2x 2 5xy + 3y 2 (viii) ( 7x + 5 ) ( 5x + 7 ) = ( 7x + 5 ) ( 5x) + ( 7x + 5 ) ( 7) = 35 x 2 + 5x + 7x + 35 = 35x x + 35

40 (xi) (2a+3b) (2a 3b) = (2a+3b) 2a + (2a+3b) ( 3b) = 4a 2 + 6ab 6ab 9b 2 = 4a 2 9b 2 (xii) (6xy 5) (6xy+5) = (6xy 5) (6xy) + (6xy 5) 5 = 36x 2 y 2 30xy + 30xy 25 = 36x 2 y 2 25 (xiii) 2 x x 7 = x x + 2 x + 3 ( 7 ) = 4 x x 14 x 21 = 4 x 2 8 x 21

41 3. Expand the following using appropriate identity: (i) (a +5) 2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = a b = 5 (a +5) 2 = a 2 +2.a.5 +b 2 = a 2 +10a +25 (ii) (2a +3) 2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = 2a b = 3 (2a +3) 2 = (2a) a.3 +b 2 = 4a a + 9 (iii) ( x + 1 x )2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = 2a b = 1 x (X + 1 x )2 = x x. 1 x + ( 1 x )2 = x x 2

42 (iv) ( 12a + 6b ) 2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = 12a and b = 6b 12a + 6b ) 2 = ( 12a) a + 6b + ( 6b) 2 = 12a ab +6b 2 = 12a ab +6b 2 = 12a ab +6b 2 (v) (π )2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = π and b = 22 (π )2 = π π ( 22 7 )2 = π π 7 = π π ( 22 7 )

43 (vi) (y 3) 2 Using (a b) 2 = a 2 2ab + b 2 a = y and b = 3 (y 3) 2 = y 2 2. y = y 2 6y + 9 (vii) (3a 2b) 2 Using (a b) 2 = a 2 2ab + b 2 a = 3a and b = 2b (3a 2b) 2 = (3a) 2 2.3a.2b + (2b) 2 = 9a 2 12ab + 4b 2 (viii) ( y 1 y )2 Using (a b) 2 = a 2 2ab + b 2 a = y and b = 1 y (y 1 y )2 = y 2 2.y. 1 y + ( 1 y )2 = y y 2

44 (ix) ( 10x 5y) 2 Using (a b) 2 = a 2 2ab + b 2 a = 10xand b = 5y) ( 10x 5y) 2 = ( 10x) x. 5y)+ ( 5y)) 2 = 10x xy + 5y 2 = 10x xy + 5y 2 = 10x xy + 5y 2 (x) (π 22 7 )2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = π and b = 22 (π 22 7 )2 = π 2 2. π ( 22 7 )2 = π 2 44 π ( 22 7 )2 = π 2 44π

45 (xi) (2x+3) (2x+5) Using (x + a) (x + b) = x 2 + x (a + b) ab we get x = 2x, a = 3 and b = 5 (2x+3) (2x+5) = (2x) 2 +2x (3 + 5) = 4x 2 +16x +15 (xii) (3x 3) (3x + 4) Using (x + a) (x + b) = x 2 + x (a + b) ab we get x = 3x, a = 3 and b = 4 (3x 3) (3x + 4) = (3x) 2 + 3x [( 3)+(4)] + (-3)4 = 9x 2 + 3x 12 = 9x 2 + 3x Expand : (i) (x + 3 ) (x 3) Using (a + b) (a b) = a 2 b 2 we get a = x, b = 3 (x + 3) (x 3) = x = x 2 9

46 (ii) (3x 5y) (3x + 5y) Using (a + b) (a b) = a 2 b 2 we get a = 3x, b = 5y (3x 5y) (3x + 5y) = (3x) 2 (5y) 2 = 9x 2 25y 2 (iii) x 3 + y 2 x 3 + y 2 Using (a + b) (a b) = a 2 b 2 we get a = x 3, b = y 2 x 3 + y 2 x 3 + y 2 = ( x 3 )2 - ( y 2 )2 (iv) (x 2 + y 2 ) (x 2 y 2 ) = x2 + y Using (a + b) (a b) = a 2 b 2 we get a = x 2, b = y 2 (x 2 + y 2 ) (x 2 y 2 ) = (x 2 ) 2 (y 2 ) 2 = x 4 y 4

47 (v) (a 2 + 4b 2 ) (a + 2b) (a - 2b) Using (a + b) (a b) = a 2 b 2 for 2 nd and 3 rd termwe get (a 2 + 4b 2 ) (a + 2b) (a - 2b) = (a 2 + 4b 2 ) [a 2 (2b) 2 ] = (a 2 + 4b 2 ) (a 2 4b 2 ) Using the above identity once again we get = (a 2 ) 2 (4b 2 ) 2 = a 4 16b 4 (vi) (x 4) (x + 4) (x 3) (x + 4) Using (a + b) (a b) = a 2 b 2 we get (x 4) (x + 4) (x 3) (x + 4) = (x ) (x ) = (x 2 16) (x 2 9) Using (a + b) (a b) = x 2 x (a + b) + ab = (x 2 ) 2 x (16+9) = x 4 25x

48 (vii) (x a) (x + a) 1 x 1 a 1 x + 1 a (x 2 a 2 ) x 2 a 2 x 2 x 1 x 2 x2 x 1 a 2 a2 x 1 + x 2 a2 x 1 a 2 1 x2 a 2 a2 x x2 a2 a2 x 2 5. Simplify the following: (i) (2x 3y) xy = (2x) 2 + (3y) 2 2.2x.3y + 12xy = 4x 2 + 9y 2-12xy +12xy = 4x 2 + 9y 2 (ii) (3m + 5n) 2 (2n) 2 = (3m) 2 + (5n) m.5n 4n 2 = 9m n mn 4n 2 = 9m mn +21n 2

49 (iii) (4a 7b) 2 (3a) 2 = (4a) 2 2.4a.7b + (7b) 2 (3a) 2 = 16a 2 56ab + 49b 2-9a 2 = 7a 2-56ab + 49b 2 (iv) (x + 1 x )2 (m + 1 m )2 = (x x. 1 x + 1 = x = x x 2)2 (m m. 1 m + 1 m 2)2 x 2 (m m 2) x 2 m m 2 = x 2 m x 2 1 m (v) (m 2 + 2n 2 ) 2 4m 2 n 2 = m 4 +2m 4.2n 2 +4n 4 4m 2 n 2 = m 4 + 4m 2 n 2 + 4n 2 4m 2 n 2 = m 4 4n 2

50 (vi) (3a 2) 2 (2a -3) 2 = (9a 2 2.3a ) (4a 2 2.3a ) = 9a 2 12a + 4 4a a 9 = 5a 2 5 =5(a 2 1)

51 EXERCISE Find the following products: (i) (x + 4) (x + 5) (x + 2) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get a = 4, b = 5 and c =2 (x + 4) (x + 5) (x + 2) = x 3 + x 2 ( ) + x ( ) = x x 2 + x ( ) +40 = x x x +40 (ii) (y + 3) (y + 2) ( y 1) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = y, a = 3, b = 2 and c = -1 (y + 3) (y + 2) (y 1) = y 3 + y 2 ( ) + y ( (-1) + (-1) (-1) = y 3 + 4y 2 + y (6 2 3) 6 = y 3 + 4y 2 + y 6

52 (iii) (a + 2) (a 3) (a + 4) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = a, a = 2, b = 3 and c = 4 (a + 2) (a 3) (a 4) = a 3 + a 2 ( ) + a [2( 3) + ( 3) ) + 2 ( 3) 4 = a 3 + 3a 2 + a ( ) 24 = a 3 + 3a 2 10a 24 (iv) (m 1) (m 2) (m 3) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = m, a = 1, b = 2 and c = 3 (m 1) (m 2) (m 3) = m 3 + m 2 ( 1 2 3) + m [( 1) ( 2) + ( 2) ( 3) + ( 3)( 1)] + ( 1) ( 2) ( 3) = m 3 + m 2 ( 6) + m [ ] 6 = m 3-6m m 6

53 (v) ( 2 + 3) ( 2+ 5) ( 2+ 7 ) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = 2, a = 3, b = 5 and c = 7 ( 2 + 3) ( 2+ 5) ( 2+ 7 ) = ( 2) 3 + ( 2) 2 3) = ) (vi) 105 x 101 x 102 We can write this as ( ) ( ) ( ) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = 100, a = 5, b = 1 and c = 2 ( ) ( ) ( ) = ( ) ( ) = (8) + 100( ) + 10 = =

54 (vii) 95 x 98 x 103 We can write this as (100-5) (100-2) ( ) Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = 100, a = -5, b = -2 and c = 3 (100 5) (100 2) ( ) = ( ) + 100( 5) ( 2) + ( 2) 3 (viii) 1.01 x 1.02 x 1.03 We can write this as ( ) ( ) ( ) + 3 (-5) + (-5) (-2) 3 = ( 4) +100 ( ) + 30 = = Using (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get x = 1, a = 0.01, b = 0.02 and c = 0.03 ( ) ( ) ( ) = ( ) + 1 [(0.01) (0.02) + (0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03) = ( ) =

55 2. Find the coefficients of x 2 and x in the following: (i) (x + 4) (x + 1) (x + 2) = x 3 + x 2 ( ) + x ( ) = x 3 + 7x x + 8 Coefficient of x 2 is 7 Coefficient of x is 14 (ii) (x 5) (x 6) (x 1) = x 3 + x 2 ( 5 6 1) + x [( 5) ( 6) + ( 6) ( 1) 1( 5)] + ( 5) ( 6) ( 1) = x 3 12x 2 + x ( ) 30 = x 3 12x x 30 Coefficient of x 2 is 12 and x is 41 (iii) (2x + 1) (2x 2) (2x 5) = (2x) 3 + (2x) 2 [1 2 5] + 2x [(1)( 2) + ( 2)( 5) + ( 5)(1)] + 1 ( 2) ( 5) = 8x 3 + 4x 2 ( 6) + 2x [ ] + 10 = 8x 3 24x 2 + 6x + 10 Coefficient of x 2 is 24 and x is 6

56 (iv) ( x 2 + 1) ( x 2 + 2) ( x 2 + 3) = ( x 2 )3 + ( x 2 )2 [ ] + x 2 [ ] = x3 8 + x2 4 (6) + x 2 ( ) + 6 = x x x + 6 Coefficient of x 2 is 3 2 and x is The length, breadth and height of a cuboid are (x +3), (x - 2) and (x -1) respectively. Find its volume. Volume of a cuboid = length x breadth x height V = (x +3) (x 2) (x 1) Using the identity (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get V = x 3 + x 2 (3 2 1) + x [3( 2) + ( 2) ( 1) + ( 1)3] + 3( 2) ( 1) = x 3-0x 2 + X ( ) + 6 = x 3-7x 2 + 6

57 4. The length, breadth and height of a metal box are cuboid are (x +5), (x 2) and (x 1) respectively. What is its volume? Volume of the metal box = length x breadth x height V = (x +5) (x 2) (x 1) Using the identity (x + a) (x + b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get V = x 3 + x 2 + ( 5 2 1) + x [5( 2) + ( 2) ( 1) + ( 1)5] + 5 ( 2) ( 1) = x 3 + 2x 2 + x [ ] + 10 = x 3 + 2x 2 13x Prove that (a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) abc [Hint: write a + b = a + b + c c, b + c = a + b + c a, c + a = a + b + c d] x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc we get L. H.S. = (a + b + c) 3 + (a + b + c) 2 ( c a b) + (a + b + c) [( c) ( a) + ( a) ( b) + ( b) ( c)] ( c) ( a) ( b) = (a + b + c) 3 -(a + b + c) 2 [(a + b + c)] +(a + b + c) (ac + ab + bc) abc = (a + b + c) 3 - (a + b + c) 3 + (a + b + c) (ac + ab + bc) abc = (a + b + c) (ac + ab + bc) abc = R. H. S.

58 6. Find the cubes of the following: (i) (2x +y) 3 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 2x, b = y (2x +y) 3 = (2x) 3 + 3(2x) 2 y + 3 (2x) y 2 +y 3 = 8x x 2 y + 6 xy 2 +y 3 (ii) (2x + 3y) 3 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 2x, b = 3y (2x + 3y) 3 = (2x) 3 + 3(2x) 2 (3y) + 3 (2x) (3y) 2 + (3y) 3 = 8x x 2 y + 54 xy y 3 (iii) (4a + 5b) 3 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 4a, b = 5b (4a + 5b) 3 = (4a) 3 + 3(4a) 2 (5b) + 3(4a)(5b) 2 + (5b) 3 = 64a a 2 b + 300ab b 3

59 (iv) ( x + x 1 )3 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = x, b = x 1 (x + x 1 )3 = x 3 + 3x 2 x + 3x ( x 1 1 )3 + ( x 1 )3 = x 3 + 3x + 3x x x 3 = x 3 + 3x x x 3 (v) 23 3 We write this as (20 + 3) 3 Using identity (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 20, b = 3 (20 + 3) 3 = (20) (20) 2 (3) + 3(20) = = (vi) 51 3 We write this as (50 + 1) 3 Using identity (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 50, b = 1 (50 + 1) 3 = (50) 3 3 x (50) 2 x 1 + 3(50) (1)

60 = = (vii) We write 101 as ( ) 3 Using identity (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 100, b = 1 ( ) 3 = = = (viii) We write 2.1 ( ) 3 Using identity (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 2, b = 0.1 ( ) 3 = x 2 2 (0.1) + 3 x 2 x (0.1) 2 + (0.1) 3 = = 9.261

61 7. Find the cubes of the following: (i) (2a 3b) 3 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 2a, b = 3b (2a 3b) 3 = (2a) 3 3 (2a) 2 (3b) + 3 (2a)(3b) 2 (3b) 3 = 8a 3 36a 2 b + 54ab 2-27b (ii) ( x 1 x )3 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = x, b = 1 x (x 1 x )3 = x 3 3x x( 1 x x )3 ( 1 x )3 = x 3 3x + 3x x 2 1 x 3 = x 3 3x x x 3 (iii) ( 3 x 2) 2 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 3 x, b = 2 ( 3 x 2) 2 = ( 3x) 3 3 ( 3x) x x = 3 3 x x x 8 =3 3 x 3 18x x 8

62 (iv) (2x 5) 3 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 2x, b = 5 (2x 5) 3 = (2x) 3 3(2x) x. 5) 2 ( 5) 3 = 8x x x 5 5 (v) 49 3 We can write 49 = 50 1 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 50, b = ) 3 = = x = = (vi) 18 3 We wrote 18 = 20 2 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 20, b = 2 (20 2) 3 = = x x4-8 =

63 = 5832 (vii) 95 3 We write 95 = Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 100, b = 5 (100 5) 3 = = = (viii) We write = (110 2) Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 110, b = -2 (110 2) 3 = (110) x = =

64 8. If x + 1 = 3, prove that x x3 + 1 = 18. x3 Given x + 1 x = 3 Cubing both sides we get (x + 1 x )3 = 3 3 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = x b = 1 x (x + 1 x )3 = (x) 3 + ( 1 x )3 + 3x. 1 x (x + 1 x ) = 27 = x x (3) = x x 3 = 27 9 = x x 3 = If p + q = 5 and pq = 6, find p 3 + q 3 (p + q) 3 = p 3 + 3pq (p + q) + q = p (5) + q = p q 3 p 3 + q 3 = p 3 + q 3 = 35

65 10. If a b = 3 and ab = 10, find a 3 b 3 Given a b = 3 and ab = 10 (a b) 3 = a 3 b 3-3ab (a b) 3 3 = a 3 b (3) 27 = a 3 b 3 90 a 3 b 3 = a 3 b 3 = If a a 2 = 20 and a3 + 1 a 3 = 30, find a + 1 a a a 3 = (a + 1 a ) (a2 + 1 a 2 - a x 1 a ) 30 = (a + 1 a ) = (20 1) 30 = (a + 1 a ) x = a + 1 a a + 1 a = 30 19

66 EXERCISE Expand the following: (i) (a + b + 2c) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = a, b = b and c = 2c (a + b + 2c) 2 = a 2 + b 2 + (2c) 2 + 2ab + 2b (2c) + 2 (2c)a = a 2 + b 2 + 4c 2 + 2ab + 4bc + 4ca (ii) (x + y + 3z) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = x, b = y and c = 3z (x + y + 3z) 2 = x 2 + y 2 + (3z) x.y + 2y(3z) + 2.(3z)x = x 2 + y 2 + 9z 2 + 2xy + 6yz + 6zx (iii) (p + q - 2r) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = p, b = q and c = -2r (p + q - 2r) 2 = p 2 + q 2 + (-2r) p.q + 2q(-2r) +2(-2r)p = p 2 + q 2 + 4r 2 + 2pq 4pr 4pr

67 (iv) ( a 2 + b 2 + c 2 )2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = a 2, b = b 2 and c = c 2 ( a 2 + b 2 + c 2 )2 =( a 2 )2 + ( b 2 )2 + ( c 2 )2 + 2 ( a 2 ) ( b 2 ( c 2 ) ( a 2 ) = a2 4 + b2 4 + c ab 2 + bc 2 + ca 2 2 ) + 2 ( b 2 ) ( c 2 ) + (v) (x 2 + y 2 + z) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = x 2, b = y 2 and c = z (x 2 + y 2 + z) 2 = (x 2 ) 2 + (y 2 ) 2 + (z) 2 + 2x 2 y 2 + 2y 2 z +2zx 2 = x 4 + y 4 + z 2 + 2x 2 y 2 + 2y 2 z +2zx 2 (vi) (m 3-1 m )2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = m, b = -3 and c = - 1 m (m 3-1 m )2 = m 2 + (-3) 2 + ( 1 m )2 + 2.m(-3) + 2(-3)(- 1 m ) + 2 (- 1 m ) m = m m 2-6m + 6 m - 2

68 = m m m 6m + 7 (vii) (-a + b c) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = a b = b c = c (-a + b c) 2 = ( a) 2 + b 2 + ( c) 2 + 2( a)b + 2b( c) +2( c)a = a 2 + b 2 + c 2 2ab 2bc +2ca (viii) (x x )2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = x b = 5 c = 1 2x (x x )2 = x ( 1 2x )2 + 2.x x + 2( 1 2x )x = x x 2 +10x + 5 x + 1 = x x 2 +10x + 5 x + 26

69 2. Simplify the following: (i) (a b + c) 2 (a b c) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get (a b + c) 2 (a b c) 2 = [ a 2 + (-b) 2 + c 2 + 2a(-b) + 2(-b)c + 2ca ] [a + (-b) 2 + ( c) 2 + 2a( b) + 2( b)( c) + 2( c)a] = a 2 + b 2 + c 2 2ab 2bc +2ca [a 2 + b 2 + c 2 2ab + 2bc 2ca] = a 2 + b 2 + c 2 2ab 2bc +2ca a 2 b 2 c 2 + 2ab 2bc +2ca = 4ac 4bc = 4c (a b) (ii) (3x + 4y + 5) 2 (x + 5y 4) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get (3x + 4y + 5) 2 (x + 5y 4) 2 = [(3x) 2 + (4y) x.4y + 2.4y (3x)] [x 2 + (5y) 2 + (-4) X.5y + 2.5y (-4) + 2 (-4). x] = 9x y xy + 40y + 30x [x y xy 40y-8x = 9x y xy + 40y + 30x - x 2-25y xy + 40y + 8x = 8x 2 9y 2 14xy + 80y + 38x + 9

70 (iii) (2m n - 3p) 2 + 4mn - 6np + 12pm Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get (2m n - 3p) 2 + 4mn - 6np + 12pm = (2m) 2 + ( n) 2 + (-3p) m( n) + 2(n)( 3p) + 2 (-3p)(2m) + 4mn 6np + 12pm = 4m 2 + n 2 + 9p 2 4mn 6np 12pm + 4mn 6np + 12pm = 4m 2 + n 2 + 9p 2 (iv) (x + 2y + 3z + r) 2 + (x + 2y + 3z r) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get a = (x + 2y) b = 3z c = r (x + 2y + 3z + r) 2 + (x + 2y + 3z r) 2 = (x + 2y) 2 + (3z) 2 + r (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y) 2 + (3z) 2 (r) (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y) = 2(x + 2y) 2 + 9z 2 + r (x + 2y)z + 6zr + 2r (x + 2y) + 9z 2 + r (x + 2y) z - 6zr 2r (x + 2y) = 2(x x.2y +4y 2 ) + 18z 2 + 2r (x + 2y)z = 2x 2 + 8xy + 8y z 2 + 2r 2 +12xz + 24 yz = 2x 2 + 8y z 2 + 2r 2 + 8xy +12xy + 24 yz

71 3. If a + b + c = 12 and a 2 + b 2 + c 2 = 50, find ab + bc + ca. Given a + b + c = 12 squaring both sides (a + b + c) 2 = 12 2 a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 144 Given a 2 + b 2 + c 2 = ab + 2bc + 2ca = 144 2(ab + bc + ca) = (ab + bc + ca) = 94 ab + bc + ca = 94 2 ab + bc + ca = If a 2 + b 2 + c 2 = 35 and ab + bc + ca = 23, find all possible values of a + b +c. Given a 2 + b 2 + c 2 = 35 and ab + bc + ca = 23, (a + b +c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = a 2 + b 2 + c 2 + 2(ab + bc + ca) = (23) = (a + b + c) 2 = 81 (a + b +c) = ± 81 a + b +c = ± 9

72 5. Express 4x + 9y + 16z + 12xy 24yz 16zx as the square of a trinomial Using and comparing the coefficient of (a +b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc +2ca we get 4x + 9y + 16z + 12xy 24yz 16zx = (2x) 2 + (3y) 2 + ( 4z 2 ) + 2.2x.3y.( 4z) + 2.( 4z) + 2.( 4z).2x = (2x + 3y 4z) 2 6. If x and y are real numbers and satisfy the equation (2x + 3y 4z) 2 + (5x y - 4) 2 = 0 find x, y. [Hint: If a, b are real numbers such that a 2 + b 2 = 0, then a = b = 0.] Given if a 2 + b 2 = 0, then a = b = 0 (2x + 3y 4z) 2 = 0 and (5x y - 4) 2 = 0 2x + 3y = 5 x 1 5x y = 4 x 3 2x + 3y = 5 2x + 3y = y = 5 15x 3y = y = 5 17x = 17 3y = 5 2 x = y = 3 x = 1 y = 3 3 = 1 x = 1 y = 1

73 EXERCISE I. If a + b + c = 0, prove the following: (i) (b + c) (b c) + a (a + 2b) = 0 Given a + b + c = 0 the a + b = -c, b + c = -a, c + a = -b we have L.H.S = (b + c) (b c) + a (a + 2b) = (-a) (b c) + a (a + b + b) = -ab + ac + a (-c + b) = -ab + ab - ac + ac = 0 = R.H.S (ii) a (a 2 bc) + b (b 2 c) + c (c 2 ab) = 0 L.H.S = a (a 2 bc) + b (b 2 c) + c (c 2 ab) = a 3 abc + b 3 - abc + c 3 abc = a 3 + b 3 + c 3 3abc We know that if a + b + c = 0 then a 3 + b 3 + c 3 = 3abc Hence we have = 3abc 3abc = 0 = R.H.S

74 (iii) a (b 2 + c 2 ) + b (c 2 + a 2 ) + c (a 2 + b 2 ) = 3abc L.H.S = a (b 2 + c 2 ) + b (c 2 + a 2 ) + c (a 2 + b 2 ) = ab 2 + ac 2 + bc 2 + ba 2 + ca 2 +cb 2 = ab 2 + ba 2 + b 2 c + bc 2 + ac 2 + a 2 c = ab (a + b) + bc (b + c) + ac (a + c) = ab ( c) + bc ( a) + ac ( b) = abc abc abc = 3abc = R.H.S [a + b + c =0, a + b = -c, b + c = -a, c + a = -b] (iv) (ab + bc + ca) 2 = a 2 b 2 + b 2 c 2 + c 2 a 2 L.H.S = (ab + bc + ca) 2 = (ab) 2 + (bc) 2 + (ca) 2 +2ab.bc + 2bc. ca + 2ca.ab = a 2 b 2 + b 2 c 2 + c 2 a 2 + 2ab 2 c + 2bc 2 a + 2ca 2 b = a 2 b 2 + b 2 c 2 + c 2 a 2 + 2abc + (0) = a 2 b 2 + b 2 c 2 + c 2 a 2 = R.H.S

75 (v) a 2 bc = b 2 ca = c 2 ab = - (ab + bc + ca) a. a bc b 2 ca c 2 ab a ( b c) bc b. b ca c. c ab b( c a) ca c( a b) ab ab ac ba bc ab ca ac bc ab (ab + bc + ca) (ab + bc + ca) (ab + bc + ca) (1). (2) (3) From equation (1) (2) and (3) a 2 bc = b 2 ca = c 2 ab = - (ab + bc + ca) (vi) 2a 2 + bc = (a b) (a c) L.H.S = 2a 2 + bc = a 2 + a 2 + bc = a 2 + a x a + bc = a 2 + a (- b c) + bc = a 2 ab ac + bc = a (a b) c (a - b) = (a b) (a c) = R.H.S

76 (vii) (a + b) (a b) + ca - cb = 0 We have a + b + c = 0 a + b = c L.H.S = (a + b) (a b) + ac cb = c (a b) + ac cd = ca + bc + ac cb = 0 = R.H.S (viii) a 2 + b 2 + c 2 = -2(ab + bc + ca) We have a + b + c = 0 Squaring we get (a + b + c) 2 = 0 a 2 + b 2 + c 2 + 2ab + 2bc +2ca = 0 a 2 + b 2 + c 2 = 2ab 2bc 2ca a 2 + b 2 + c 2 = 2(ab + bc + ca) Hence the proof

77 2. Suppose a, b, c are non-zero real numbers such that a + b + c = 0, Prove the following: (i) a 2 + b 2 + c 2 bc ca ab = 3 L.H.S = a2 + b 2 + c 2 bc ca ab = a2.a+b 2.b +c 2.c abc = a3 +b 3 +c 3 abc. (1) We have a + b + c = 0, a + b = c Cubing we get (a + b) 3 = ( c) 3 a 3 + b 3 + 3ab + (a + b) = c 3 a 3 + b 3 3ab = c 3 a 3 + b 3 + c 3 = 3abc. (2) Substituting (2) in (1) L.H.S = 3abc abc = 3

78 (ii) ( a+b c + b+c a + c+a b ) ( b c+a + c a+b + a b+c ) Whenever b + c 0, c + a 0, a + b 0 We have a + b + c =0 a + b = c b + c = a c + a = b L.H.S = ( a +b c + b+c a + c+a b ) ( b c+a + c a +b + a b +c ) = ( c c + a a + a b ) ( b b + c c + a a ) = (-1-1-1) (-1-1-1) = (-3) (-3) = 9 = R.H.S (iii) a 2 + b2 + c2 2a 2 +bc 2b 2 ca 2c 2 = 1, provided the denominators do not become ab 0. L.H.S = a 2 + b 2 + c 2 2a 2 +bc 2b 2 ca 2c 2 ab = a 2 a b (a c) + b 2 + b c (b a) c 2 c a (c b) = a 2 a b (a c) - b 2 + a b (b c) c 2 a c (b c)

79 = a2 b c b 2 a c + c 2 a b a b (b c)(a c) = a2 b a 2 c b 2 a + b 2 c + c 2 a c 2 b ab b 2 ac +bc (a c) = a 2 b a 2 c b 2 a + b 2 c + c 2 a c 2 b a 2 b ab 2 a 2 c + abc abc + b 2 c + ac 2 = 1 = R.H.S 3. If a + b + c = 0, prove that b 2 4ac is a square. We have a + b + c = 0 b = - (a + c) Squaring on both sides b 2 = [- (a + c)] 2 = (a + c) 2 b 2 = a 2 + c 2 + 2ac Subtracting 4ac on both sides b 2 4ac = a 2 + c 2 + 2ac 4ac = a 2-2ac + c 2 b 2 4ac = (a - c) 2 We find that b 2 4ac is the square of (a c)

80 4. If a, b, c are real numbers such that a + b + c = 2s, prove the following: (i) s (s a) + s (s - b) + s (s c) = s 2 L.H.S. = s (s a) + s (s - b) + s (s c) = s 2 as + s 2 bs + s 2 cs = 3s 2 as bs cs = 3s 2 s (a + b + c) = 3s 2 s (2s) (a + b + c = 2s) = 3s 2 2s 2 = s 2 = R.H.S. (ii) s 2 (s a) 2 + s (s - b) 2 + s (s c) 2 = a 2 + b 2 + c 2 L.H.S. = s 2 (s a) 2 + s (s - b) 2 + s (s c) 2 = s 2 + s 2 + a 2 2sa + s 2 + b 2 + s 2 + c 2 2as 2bs 2cs = 4s 2 + a 2 + b 2 + c 2 2s 2bs 2cs = 4s 2 + a 2 + b 2 + c 2 2s (a + b + c) = 4s 2 + a 2 + b 2 + c 2 2as (2s) (a + b + c = 2s) = 4s 2 + a 2 + b 2 + c 2 4s 2 = a 2 + b 2 + c 2 = R.H.S.

81 (iii) (s a) (s b) + (s b) (s c) + (s c) (s a) + s 2 = ab + bc + ca L.H.S. = (s a) (s b) + (s b) (s c) + (s c) (s a) + s 2 = s 2 as bs + ab + s 2 bs cs + bc + s 2 cs as + ac + s 2 = 4s 2 2 as 2bs 2cs + ab + bc +ca = 4s 2 2s (a + b + c) + ab + bc +ca = 4s 2 2s (2s) + ab + bc +ca = 4s 2 4s 2 + ab + bc +ca (a + b + c = 2s) = ab + bc +ca = R.H.S. (iv) a 2 b 2 c 2 + 2bc = 4 (s b) ( s c) L.H.S = a 2 b 2 c 2 + 2bc = a 2 (b 2 + c 2 2bc) = a 2 (b c) 2 = (a + b c) [a (b c)] = (a + b c) (a + b c) = (2s c c) (2s b b) = (2s 2c) (2s 2b) = 2(s c) (2) (s b) = 4 (s c) (s b) = R.H.S.

82 5. If a, b, c are real numbers, a + b + c =2s and s a 0, s b 0, s c 0, Prove that L.H.S = a + b + (s a) (s b) a (s a ) + b (s b) + c (s c) + 2 = c (s c) + 2 s c abc s b (s c) = a s b s c + b s a s c + c s a s b + 2(s a)(s b)(s c) s a s b (S c) = [a s 2 bs cs +bc +b s 2 as cs+ac +c s 2 as bs +ab +2 s 2 as bs +ab s c ] s a s b (S c) = as 2 abs acs +abc +b s 2 abc bcs +abc +cs 2 acs bcs +abc +2(s 3 as 2 b s 2 +abs cs 2 +acs +bcs abc ] s a s b (S c) = [s 2 a+b+c 2abc 2acs 2bcs 3abc + 2s 3 2 as 2 2bs 2 + 2abs 2cs 2 + 2acs + 2bcs 2abc ] s a s b (S c) = [s2 2s + abc +2s 3 2s 2 a+b+c ] (s a) s b (S c) = [2s3 + abc +2s 3 2s 2 2s ] (s a) s b (S c) = 4s3 + abc 4s 3 (s a ) s b (S c) = abc (s a ) s b (S c) = R.H.S.

83 6. If a + b + c = 0, prove that a 2 bc = b 2 ca = c 2 ab = (a2 + b 2 +c 2 ) We have a + b + c = 0 Squaring will get (a + b + c) 2 = 0 a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 0 a 2 + b 2 + c 2 + 2b (a + c) + 2ca = 0 a 2 + b 2 + c 2 + 2b (-b) + 2ca = 0 2 a 2 + b 2 + c 2 = 2b 2 2ca (a + c = -b) a 2 + b 2 + c 2 = 2 (b 2 ca) a 2 + b 2 +c 2 2 = b 2 ca (1) IIIly (a + b + c) 2 = 0 a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 0 a 2 + b 2 + c 2 + 2ab + 2c (b + a) = 0 a 2 + b 2 + c 2 + 2ab + 2c (-c) = 0 (a + c = -c) a 2 + b 2 + c 2 = 2 (c 2 ab) a 2 + b 2 +c 2 2 = (c 2 ab) (2)

84 Also a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 0 a 2 + b 2 + c 2 + 2a (b + c) + 2bc = 0 a 2 + b 2 + c 2 + 2a (-a) + 2bc = 0 a 2 + b 2 + c 2 = 2 (a 2 bc)..(3) From (1), (2) and (3) we get, a 2 bc = b 2 ca = c 2 ab = (a2 + b 2 +c 2 ) 2 7. If 2(a 2 + b 2 ) = (a + b) 2, prove that a = b. 2a 2 + 2b 2 = a 2 + b 2 + 2ab 2a 2 + 2b 2 a 2 b 2 2ab = 0 a 2 + b 2 2ab = 0 (a b) 2 = 0 a b =0 a = b

85 8. If x 2 3x + 1 = 0, prove that x = 7. x2 Given x 2 3x + 1 = 0 x = 3x x + 1 x = 3 (dividing both sides by x) Squaring both sides we get (x + 1 x )2 = 3 2 = x x 2 + 2x. 1 x = 9 x x = 9 x x 2 = 9 2 = 7

86 ADDITIONAL PROBLEMS ON MULTIPLICATION OF POLYNOMIALS I. Find the following products: (i) (2a + 3b) (4a 2 6ab + 9b 2 ) = 8a a 2 b 12ab 2 18ab ab b 3 = 8a b 3 (ii) (3x + 4y) (9x 2 12xy +16y 2 ) = 27x x 2 y - 36x 2 y - 48xy xy y 3 = 27x y 3 (iii) (5x 2y) (25x xy + 4y 2 ) = 125x 3 50x 2 y + 50x 2 y 20xy xy 2 8y 3 = 125x 3 8y 3 (iv) (a 3 2) (a 3 + 2a 3 + 4) = a 9 2a 6 + 2a 6 4a 3 + 4a 3 8 = a 9 8

87 2. By which factor should the following get multiplied to be in the form a 3 + b 3. (i) (2x + 1) We have a 3 + b 3 = (a + b) (a 2 ab +b 2 ) a = 2x b = 1 (2x) = (2x + 1) [(2x) 2 2x ] 8x + 1 = (2x + 1) (4x 2 2x + 1) We have to multiply (2x + 1) with (4x 2 2x + 1) (ii) 4x 2 6x + 9 We have a 3 + b 3 = (a + b) (a 2 ab +b 2 ) a = 2x b = 3 (2x) = (2x + 3) (4x 2 6x + 9) (2x + 3) should be multiplied (iii) 9a 2 15a +25 We have a 3 + b 3 = (a + b) (a 2 ab +b 2 ) a = 3a b = 5 (3a) = (3a + 5) (9a 15a + 25) Hence (3a + 5) should be multiplied.

88 (iv) 4a + 3 We have a 3 + b 3 = (a + b) (a 2 ab +b 2 ) a = 4a b = 3 (4a) = (4a + 3) [(4a) 2 4a ) = (4a + 3) (16a 2 12a + 9) Hence (16a 2 12a + 9) should be multiplied. 3. By which factor should the following get multiplied to be in the form a 3 - b 3. (i) (5a 3) We have a 3 - b 3 = (a - b) (a 2 + ab +b 2 ) a = 5a b = 3 (5a) = (5a 3 ) [(5a 2 ) + (5a) ] 125a 3 27 = (5a 3) (25a a + 9) (25a a + 9) should be multiplied. (ii) `16a a + 25 We have a 3 - b 3 = (a - b) (a 2 + ab +b 2 ) a = 4a b = 5 (4a) = (4a 5 ) (4a a + 25) (4a 5) should be multiplied.

89 (iii) 3x 2y We have a 3 - b 3 = (a - b) (a 2 + ab +b 2 ) a = 3x b = 2y (3x) 3 (2y) 3 = (3x 2y) [(3x) 2 + 3x.2y + (2y) 2 ] 27x 3 8y 3 = (3x 2y) (9x 2 + 6xy + 4y 2 ) (9x 2 + 6xy + 4y 2 ) should be multiplied. (iv) 16x 2 20x + 25 We have a 3 - b 3 = (a - b) (a 2 + ab +b 2 ) a = 4x b = 5 (4x) = (4x 5 ) (4x x + 25) (4x 5) should be multiplied. 4. Use the appropriate identity to compute the following: (i) (103) 2 We write 103 = Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 100 b = 3 ( ) 2 = = = 10609

90 (ii) (107) 2 We write 107 = Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 100 b = 7 ( ) 2 = = = (iii) ( )2 We write = 50.5 = Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 50 b = 0.5 ( ) 2 = x 50(0.5) + (0.5) 2 = (0.5) = = (iv) (998) 2 We write 998 = Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 1000 b = -2 (1000-2) 2 =

91 = = (v) 107 x 93 We write 107 = and 93 = Using (a + b) (a b) = a 2 + b 2 We get a = 100 b = 7 ( ) (100 7) = = = 9951 (vi) 1008 x 992 We write 1008 = and 992 = Using (a + b) (a b) = a 2 + b 2 We get a = 1000 b = 8 ( ) (1000 8) = = = 99936

92 (vii) (101) 3 We write 101 = Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 We get a = 100 b = 1 ( ) 3 = (100) (100) = = (viii) (1002) 3 We write 1002 = Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 We get a = 1000 b = 2 ( ) 3 = (1000) (1000) = =

93 5. If x + y = 7 and xy = 12, find x 2 + y 2 and x 2 + y 2. Given x + y = 7 and xy = 12 We take x + y = 7 Squaring both sides (x + y) 2 = 7 2 x 2 + y 2 + 2xy = 49 x 2 + y = 49 x 2 + y 2 = x 2 + y 2 = 25 IIIly x + y = 7 Cubing both sides (x + y) 3 = 7 3 x 3 + y 3 + 3xy(x + y) = 343 x 3 + y 3 + (3 x 12 x 7) = 343 x 3 + y 3 = x 3 + y 3 = 91

94 6. If x + 1 x = 3, find x2 + 1 x 2 and x4 + 1 x 4 x + 1 x = 3 Squaring both sides (x + 1 x )2 = 3 2 x x x. 1 x = 9 x x 2 = 9 2 x x 2 = 7 Squaring again (x x 2 )2 = 7 2 x x x2. 1 x 2 = 49 x x 4 = 49 2 x = 74 x 4 7. If a b = 2 and ab = 15, find a 3 b 3. a b = 2 Cubing both sides (a b) 3 = 2 3 a 3 b 3 3ab (a b) = 8 a 3 b 3 3 x 15(2) = 8

95 a 3 b 3 = a 3 b 3 = If a + b + c = 2s then prove that (s a) 3 + (s b) 3 + 3c (s a) (s b) = c 3 Let s a = x and s b = y Now L.H.S. will be x + y x 3 + y 3 + 3cxy s a + s b x 3 + y 3 + 3(x + y) xy (x + y = c) 2s a - b (x + y) 3 a + b + c a b c 3 c 9. If a + b + c = 2s, prove that 16s (s a) (s b) (s c) = 2a 2 b 2 + 2b 2 c 2 + 2c 2 a 2 a 2 a 4 b 4 c 4 Given a + b + c = 2s L.H.S. = 16s (s a) (s b) (s c) = 2s 2(s a) 2(s b) 2(s c) = 2s (2s 2a) (2s 2b) (2s 2c) = (a + b + c) (a + b + c 2a) (a + b + c 2b) (a + b + c 2c)

96 = (a + b + c) (b + c a) (a + c b) (a + b c) = (ab + b 2 + bc + ac + bc + c 2 a 2 ab ac) = (a 2 + ac ab ab + bc b 2 ac c 2 + bc) = (b 2 + c 2 a 2 + bc ) (a 2 b 2 c + 2bc) = a 2 b 2 + a 2 c 2 a 4 + 2a 2 bc b 4 b 2 c 2 + a 2 b 2 2b 3 c c 2 b 2 c 4 + a 2 c 2-2bc 3 + 2b 3 c + 2bc 3 2a 3 bc + 4 b 2 c 2 = 2a 2 b b 2 c 2 + 2a 2 c 2 a 4 b 4 c 4 = R.H.S. 10. If a 2 + b 2 = c 2, prove that (a + b + c) (b + c a) (c + a b) (a + b c) = 4a 2 b 2 L.H.S = (a + b + c) (b + c a) (c + a b) (a + b c) Re arranging (a + b + c) (a + b c) (c + b a) c (b a) (a + b) 2 c 2 c 2 (b a 2 ) a 2 + b 2 + 2ab c 2 c 2 (a 2 + b 2 2ab) (a 2 + b 2 ) + (2ab c 2 ) c 2 (a 2 + b 2 ) + 2ab (c 2 + 2ab c 2 ) c 2 c 2 + 2ab (2ab) (2ab) 4a 2 b 2

97 11. If x + y = a and xy = b, prove that (1 + x 2 ) (1 + y 2 ) = a 2 + (1 + b 2 ). ` L.H.S =(1 + x 2 ) (1 + y 2 ) = 1 + x 2 + y 2 + x 2 y 2 Add and subs tract 2xy = 1 + x 2 + y 2 + 2xy 2xy + x 2 y 2 = 1 + (x + y) 2 2xy + (xy) 2 = 1 + a 2 2b + b 2 [x + y = a, xy = b] = a 2 + (1 2b + b 2 ) = a 2 + (1 b) 2 = R.H.S [(a - b) 2 = a 2 2ab + b 2 ] 12. If x 1 = 4, show that x x3 + 6x = 184. x 2 x3 Given x 1 x = 4 Squaring both sides (x 1 x )2 = 4 2 x x 2-2x. 1 x = 16 x x 2 = x x 2 = 18..(1)

98 Again take x 1 x = 4 Cubing on both sides (x 1 x )3 = 4 3 x 3 1 x 3 3x. 1 x (x 1 x ) = 64 x 3 1 3(4) = 64 x3 x 3 1 = x3 x 3 1 x 3 = 76..(2) Now consider L.H.S x 3 + 6x x 2-1 x 3 = x x 3 (x2 + 1 ) x 2 = x 18 = = 184 R.H.S

99 13. If xy(x + y) = 1, prove that x 3 y 3 x3 y 3 = 3 Given xy(x + y) = 1 By dividing the equation by xy 1 x + y = 1 xy Cubing the both sides (x + y) 3 = ( 1 xy )3 x 3 + y 3 + 3xy(x + y) = 1 x 3 + y 3 + 3(1) = 1 x 3 y 3 1 x 3 y 3 x3 + y 3 = 3 x 3 y 3 [xy (x + y) = 1]

100 14. Suppose a, b, are the sides of a triangle such that 2s = a + b + c. Prove that a2 b 2 + 2bc c 2 = s b (s c) b 2 + 2bc + c 2 a 2 s( s a) L.H.S = a2 b 2 + 2bc c 2 b 2 + 2bc + c 2 a 2 = a2 ( b 2 2bc + c 2 ) (b 2 + 2bc + c 2 ) a 2 = a2 (b c) 2 ( b + c 2 ) a 2 = a + b c [a b c ] [ b + c + a ] [ b + c a ] [a2 b 2 = (a + b) (a b)] = = = a + b c (a b+c) a + b + c (b +c a) a + b + c c c (a + b + c b b ) a + b + c (a + b + c a a ) 2s 2c (2s 2b) 2s (2s 2a) = 4 s c (s b ) 4s (s a) = s b (s c) s (s a ) = R.H.S

101 15. Suppose a, b, are the sides of a triangle such that 2s = a + b + c. Prove that 1 s a s b s c 1 s = abc s s a s b (s c) `L.H.S. = 1 s a + 1 s b + 1 s c 1 s = s s b s c + s s a s c + s s a s b [ s a s b s c ] s s a s b (S c) = s s2 bs cs+bc +s s 2 as cs+ac +s s 2 as bs +ab s s a s b (S c) = s3 bs 2 cs 2 + bcs + s 3 a s 2 cs 2 + acs + s 3 cs 2 bs 2 + abs s s a s b (S c) = 3s 3 2as 2 2b s 2 2bc + abs + bcs + acs s 3 + a s 2 + bs 2 abs + cs 2 acs bcs +abc s s a s b (S c) = [2s 3 as 2 bs 2 cs 2 + abc ] 1 s(s a ) s b (S c) = [2s 3 s 2 a + b + c + abc ] 1 s(s a) s b (S c) = [2s 3 s 2 2s + abc ] 1 s(s a) s b (S c) (a + b + c = 2s) = 2 s3 2s 3 + abc s(s a) s b (S c) = abc s(s a) s b (S c) = RHS

102 EXTRA QUESTIONS I. Expand using appropriate identity: (i) (3a + 5b) 2 Using (a + b) 2 = a 2 + 2ab + b 2 we get a = 3a b = 5b (3a + 5b) 2 = (3a) a.5b + (5b) 2 = 9a ab + 25b 2 (ii) ( 1 2 x y)2 Using (a + b) 2 = a 2 + 2ab + b 2 we get a = 1 2 x b = 2 3 y ( 1 2 x y)2 = ( 1 2 x) x 2. 2y 3 + ( 2y 3 )2 (iii) (2a + 3b + 4c) 2 = x y2 xy Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca we get a = 2a b = 3b c = 4c (2a + 3b + 4c) 2 = (2a) 2 + (3b) 2 + (4c) a.3b + 2.3b.4c + 2.4c2a = 4a 2 + 9b c ab + 24bc + 16ca

103 (iv) (4a 3b) 2 Using (a b) 2 = a 2 2ab + b 2 we get a = 4a b = 3b (4a 3b) 2 = (4a) 2 2.4a.3b + (3b) 2 = 16a 2 24ab + 9b 2 (v) (3a 2b + 5c) 2 Using (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca we get a = 3a b = -2b c = 5c (3a 2b + 5c) 2 = (3a) 2 + ( 2b) 2 + (5c) 2 + 2(3a) ( 2b) + 2( 2b) (5c) + 2(5c) (3a) = 9a 2 + 4b c 2 12ab 20bc + 30c II. If (x 1 ) = 3, find the valve of 2x (i) x x 2 (ii) x x 4 (iii) x3 1 8x 3 (i) (x 1 2x ) = 3 Squaring on both sides (x 1 2x )2 = 3 2 x x 2 2.x. 1 2x = 9 x x 2 = = 10

104 (ii) x x 2 = 10 Squaring on both sides (x x 2 )2 = 10 2 x 4 + x x 4 2.x2 x 1 = 100 4x x = x 4 + = = x (iii) (x 1 2x ) = 3 Cubing on both sides (x 1 2x )3 = 3 3 (x 3 1 8x 3-3.x x 1 2x = (x 1 2x ) = 27 = x 3 1 = 3 (3) = 27 8x 3 2 x 3 1 = = = 63 8x

105 3. If a = 14, find the values of a2 (i) a + 1 a (ii) a 1 a (iii) a 2 1 a 2 (i) a a 2 = 14 Additing 2 on both sides (a ) = a2 (a a 2)2 = 16 a + 1 a = 12 = ±4 (ii) a a 2 = 14 Subtracting 2 on both sides (a ) = 14 2 a2 (a 1 a )2 = 12 (a 1 a ) = 12 = ±2 3 (iii) a 2 1 a 2 = (a 1 a ) (a 1 a ) = (±4) (±2 3) = (±8 3)

106 4. If (3a + 4b) = 16 and ab = 4 find the value of 9a b 2 We have (3a + 4b) = 16 Squaring on both sides (3a + 4b) 2 = a a.4b + 16b 2 =256 9a b 2 = ab = x 4 = = If a 2 4a 1 = 0 2a 0, find the values of (i) (a + 1 a ) (ii) a 1 a We have a 2 4a 1 = 0 a 2 4a = 1 Adding 4 on both sides a 2 4a + 4 = (a + 2) 2 = 5 a + 2 = 5 1 a = a = 5 2

107 (i) a + 1 a = = ( 5 2) = ( 5 2) = = (ii) a 1 a = = ( 5 2) = = = = 4(2 5 ) (2 5 ) = 4

108 6. If a + 2b + 3c = 0, show that a 3 + 8b 3 +27c 3 18abc We have a + 2b + 3c = 0 a + 2b = 3c Cubing on both sides (a + 2b) 3 = ( 3c) 3 a 3 + 8b 3 + 3a (2b) (a+2b) = -27c 3 a 3 + 8b 3 + 6ab (-3c) = -27c 3 a 3 + 8b c 3 = 18abc 7. If a b = b c prove that (a + b + c) (a b + c) = a2 + b 2 + c 2 [Hint: Let a = b = k. then b = ck, a = bk, b = ck and a = b c ck2 ] L.H.S = (a + b + c) (a b + c) = (ck 2 + ck + c) (ck 2 - ck + c) = c (k 2 + k + 1) c (k 2 - k + 1) = c 2 (k k) (k k) = c 2 [(k 2 1) 2 k] = c 2 [k 4 + 2k 2 1 k 2 ] = c 2 [k 4 + k 2 + 1]. (1)

109 R.H.S = a 2 + b 2 + c 2 = (ck 2 ) 2 + (ck) 2 + c 2 = c 2 k 4 + c 2 k 2 + c 2 = c 2 (k 4 + k 2 + 1).. (2) From (1) and (2) we get LHS = RHS 8. Evaluate (i) (968) 2 (32) 2 (ii) (98.7) 2 (1.3) 2 (i) (968) 2 (32) 2 Using (a 2 b 2 ) = (a + b) (a b) we get a = 968 and b = 32 (968) 2 (32) 2 = ( ) (968 b) = 1000 x 936 = (ii) (98.7) 2 (1.3) 2 Using (a 2 b 2 ) = (a + b) (a b) we get a = 98.7 and b = 1.3 (98.7) 2 (1.3) 2 = ( ) ( ) = 100 x 97.4 = 9740

110 9. If x + y = 8 and xy = 12, find x 4 + y 4. Take x + y = 8 Squares we get (x + y) 2 = 8 2 x 2 + y 2 + 2xy = 64 x 2 + y x 12 = 64 x 2 + y 2 = x 2 + y 2 = 40 Squaring both sides we get (x 2 + y 2 ) 2 = 40 2 x 4 + y 4 + 2x 2 y 2 = 1600 x 4 - y 4 + 2(xy) 2 = 1600 x 4 + y 4 + 2(12) 2 = 1600 x 4 + y x 144 = 1600 x 4 + y 4 = x 4 + y 4 = If a + b = 8 and ab = 15, find a 3 + b 3. Take a + b = 8 Cubing both sides

111 (a + b) 3 = 8 3 a 3 + b 3 + 3ab (a + b) = 512 a 3 + b x 15(8) = 512 a 3 + b = 512 a 3 + b 3 = a 3 + b 3 = If x 1 x = 3, find x3 1 x 3 Take x 1 x = 3 Cubing on both sides (x 1 x )3 = 3 3 x 3 + ( 1 x )3 + 3x + 1 x (x 1 x ) = 27 x (3) = 27 x3 x x 3 = 27 9 x x 3 = If x 1 x = 5, find x3 1 x 3 Take x 1 x = 5 Cubing on both sides (x 1 x )3 = 5 3

112 x 3 ( 1 x )3 3x + 1 x (x 1 x ) = 125 x 3 1 3(5) = 125 x3 x = 125 x3 x 3 1 = x3 x 3 1 x 3 = Suppose x, y, z are non-zero real numbers such that x + y + z = 0 (x + y) 2 xy + (y + z)2 yz + (z + x)2 zx = 3. We have x + y = -z, y + z = -x, and z +x = -y LHS = (x + y)2 xy + (y + z)2 yz + (z + x)2 zx = ( z)2 xy + ( x)2 zy + ( y)2 zx = z2 + x2 + y 2 xy xy xy = z3 + x 3 + y 3 xyz (if x + y + z = 0 then x 3 + y 3 + z 3 = 3xyz) = 3 xyz xyz = 3 RHS

113 14. If a + b + c = 2s prove that s(s a) + (s b) + (s c) = s 2 LHS = s(s a) + (s b) + (s c) s 2 sa + s 2 sb + s 2 - sc 3s 2 s (a + b + c) 3s 2 s (2s) s 2 = RHS 15. If x + 1 = 3, find x x2 + 1 = 3 5 x2 x + 1 x = 3 Squaring on both sides (x + 1 x )2 = 3 2 x x x 1 x =9 x x 2 = 9 2 x x 2 = 7..(1) Now (x 2 1 x )2 = (x 1 x )2 4x x 1 x (x 1 x )2 = (x 1 x )2 = 5 x 1 = 5 (2) x

114 Now x 2 12 x = (x + 1 x ) (x 1 x ) From (1) and (2) x = 3 5 = R.H.S x2 Important formulae 1. (a + b) 2 = a 2 + 2ab + b 2 2. (a b) 2 = a 2 2ab + b 2 3. (a + b) (a b) = a 2 b 2 4. (x + a) (x + b) = x 2 + x (a + b) + ab 5. (x + a) (x+ b) (x + c) = x 3 + x 2 (a + b + c) + x (ab + bc + ca) + abc 6. (a + b) 3 = a 3 + b 3 + 3ab (a + b) 7. (a + b) 3 = a 3 b 3 3ab (a b) 8. a 3 + b 3 = (a + b) (a 2 ab + b 2 ) 9. a 3 b 3 = ( a b) (a 2 + ab + b 2 ) 10. (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca

115 UNIT-1 POLYGONS EXERCISE In each of the following polygons find in degrees the sum of the interior angles and the sum of exterior angles. (i) Hexagon (ii) Octagon (iii) pentagon (iv) nonagon (v) decagon Solution: (i) Hexagon Numbers of sides, n = 6 Sum of the interior angles = (2n 4) 90 = (2 x 6 4) x 90 = (12 4) 90 = 8 x 90 = 720 Sum of the exterior angles = 360 (ii) Octagon Numbers of sides, n = 8 Sum of the interior angles = (2n 4) 90 = (2 x 8 4) x 90 = (16 4) 90 = 12 x 90 = 1080 Sum of the exterior angles = 360

116 (iii) Pentagon Numbers of sides, n = 5 Sum of the interior angles = (2n 4) 90 = (2 x 5 4) x 90 = (10 4) 90 = 6 x 90 = 540 Sum of the exterior angles = 360 (iv) Nonagon Numbers of sides, n = 9 Sum of the interior angles = (2n 4) 90 = (2 x 9 4) x 90 = (18 4) 90 = 14 x 90 = 1260 Sum of the exterior angles = 360 (v) Decagon Numbers of sides, n = 10 Sum of the interior angles = (2n 4) 90 = (2 x 10 4) x 90 = (20 4) 90 = 16 x 90 = 1440 Sum of the exterior angles = 360