UNIT-1 SQUARE ROOT EXERCISE 1.1.1 1. Find the square root of the following numbers by the factorization method (i) 82944 2 10 x 3 4 = (2 5 ) 2 x (3 2 ) 2 2 82944 2 41472 2 20736 2 10368 2 5184 2 2592 2 1296 2 648 2 324 2 162 3 81 3 27 3 9 3 = 82944 = (2 5 ) 2 (3 2 ) 2 = 2 5 x 3 2 = 288
(ii) 155236 2 155236 2 77618 197 38809 197 197 1 = 22 X 197 2 = 155236 = (2 197) 2 = 2 X 197 = 394 (iii) 19881 3 19881 3 6627 47 2209 47 = 19881 = 3 2 47 2 = 3 x 47 = 141
2. Find the square root of the following numbers. (i) 184.96 184.96 x 100 100 = 18496 100 = (23 ) 2 17 2 10 2 2 18496 2 9248 2 4624 2 2312 2 1156 2 578 289 17 289 17 184.96 = (2 3 ) 2 17 2 10 2 = 8x17 10 = 136 10 = 13.6
(ii) 19.5364 19.5364 = 195364 10000 2 195364 2 97682 221 48841 19.5364 221 = 195364 10000 = ( 2 221 ) 2 100 2 = 442 100 = 4.42
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-2 10 2 = 100 digits-3 99 2 = 9801 digits-4 100 2 = 10000 digits-5 990 2 = 998001 digits-6 1000 2 = 1000000 digits-7 9999 2 = 99980001 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.
EXERCISE 1.1.2 1. Find the square root of the following number by division method: (i) 5329 73 7 5329 7 49 143 429 429 0 5329 = 73 (ii) 18769 137 1 18769 1 1 23 087 3 69 267 1869 1869 0 18769 = 137
(iii) 28224 168 1 2 82 24 1 1 26 182 6 156 328 2624 2624 0 28224 = 168 (iv) 186624 4 3 2 4 186624 4 16 83 266 3 249 862 1724 1724 0 186624 = 432
2. Find the least number to be added to get a perfect square. (i) 6200 78 7 6200 7 49 148 1300 1184 116 6200 > 78 2 Therefore we find 79 2 79 2 = 6241 6241 6200 41 is the least number to be added to 41 get a perfect square. (ii) 12675 112 1 1 26 75 1 1 21 26 1 21 222 575 444 131
113 2 = 12769 12769 12675 94 94 is the least number to be added to 12675 to make it a perfect square. (iii) 88417 297 2 8 84 17 2 4 49 484 9 441 587 4317 4109 208 298 2 = 88804 88804 88417 387 387 is the least number to be added to 88417 to make it a perfect square.
(iv) 123456 3 5 1 3 12 34 56 3 9 65 334 5 325 701 956 701 255 123456 lies between 351 2 and 352 2 351 2 = 123201 352 2 = 123904 123904 123456 448 703 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) 1234 3 5 3 12 34 3 9 65 334 325 9 9 is the least number to be the least number to be subtracted to make 1234 a perfect square.
(ii) 4321 6 5 6 43 12 6 36 125 721 525 196 196 should be the least number to be subtracted to make 4321 a perfect square. (iii) 34567 1 8 5 1 3 45 67 1 1 28 245 8 224 365 2167 1825 342 342 should be the least number to be subtracted to make 34567 a perfect square.
(iv) 109876 3 3 1 3 10 98 76 3 9 63 198 3 189 661 976 661 315 315 should be the least number to be subtracted to make 109876 a perfect square. 4. Find the two consecutive perfect square numbers between which the following number occurs. (i) 4567 6 7 6 45 67 6 36 127 967 889 78 The number 4567 lies between 67 2 and 68 2. 67 2 = 4489 68 2 = 4624
(ii) 56789 2 3 8 2 5 67 89 2 4 43 167 3 129 468 3889 3744 145 56644 lies between the squares of 238 2 and 239 2. (iii) 88888 2 9 8 238 2 = 56644 239 2 = 57121 2 8 88 88 2 4 498 488 441 4788 3984 804 88888 lies between the squares of 298 2 and 299 2. 298 2 = 88804 299 2 = 89401
(iv) 123456 3 5 1 3 12 34 56 3 9 65 334 5 325 701 956 701 255 123456 lies between 351 2 and 352 2 351 2 = 123201 352 2 = 123904 5. 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 = 12180 sq. m. Total area = 23940 sq. m.
1 5 4 1 2 39 40 1 1 25 139 5 125 304 1440 1216 224 The area of the new square plot is 154 2 = 23715 sq. m. The person would have to loose area of 23940 23716 224 sq. m.
EXERCISE 1.1.3 1. How many digits are there after the decimal points in the following? (i) (3.16) 2 (3.16) 2 = 9.9856 Ans: 4 (ii) (1.234) 2 (1.234) 2 = 1.5227556 Ans: 6 (iii) (0.0023) 2 (0.0023) 2 = 0.00000529 Ans: 8 (iv) (1.001) 2 (1.001) 2 = 1.002001 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) 84.8241 84.8241 = 9.21 Ans: 3
(ii) 0.085849 (iii) 1.844164 0.085849 = 0.293 Ans: 3 1.844164 = 1.358 Ans: 4 (iv) 0.0089510521 0.089510521= 0.09461 Ans: 5 3. Find the square root of the following number using division method. (i) 651.7809 2 5. 5 3 2 6 51. 78 09 2 4 45 251 5 225 505 2678 2525 5103 15309 15309 0 651.7809 = 25.53
(ii) 0.431649 0.6 5 7 0 0.43 16 49 0 0 06 043 6 36 125 716 5 625 1307 9149 9149 0 0.431649 = 0.657 1 (iii) 95.4529 9. 7 7 9 95. 45 29 9 81 187 1445 7 1309 1947 13629 5 13629 0 95.5849 = 9.77
(iv) 73.393489 8. 5 6 7 8 73. 39 34 89 8 64 165 939 5 825 1706 11434 6 10236 17127 119889 119889 0 73.393489 = 8.567 4. A square garden has an are 24686.6944m 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 24686.6944m 2 Length of a side of the garden the is 24686.6644 in 157.12m.
If a trench of one meter is dug along the boundary inside the garden the length of a side of the garden will be (157.12-2) m. = 155.12m. 1z zxdrjhzkgfjh 1m 157.12 1222 The area of the remaining garden is (155.12) = 24062.2144m 2
EXERCISE 1.1.4 1. Round off the following numbers to 3 decimal places. 1. 1.5678 1.568 2. 2.84671 2.847 3. 14.56789 14.568 4. 12.987564 12.988 5. 3.3333567 3.333 2. Find the square root of the following numbers correct to 3 decimal places. (i) 12 3. 4 6 4 1 3 12. 00 00 00 00 3 9 64 300 4 256 686 4400 6 4115 6924 28400 4 27696 69281 80400 69281 11119 12 = 3.4641 3.464
(ii) 1.8 1.3 4 1 6 1 1.80 00 00 00 1 1 23 080 3 69 264 1100 4 1056 2681 4400 1 2681 26826 171900 160956 10644 1.8 = 1.3416 1.3416 = 1.342 (iii) 133 1 1. 5 3 2 5 1 1 33. 00 00 00 00 1 1 21 033 1 21 225 1200 5 1125 2303 7500 3 6909 23062 59100 2 46124 230645 1297600 1153225 144375 133 = 11.5325 11.533
(iv) 12.34 3. 5 1 2 8 3 12.34 00 00 00 3 9 65 334 5 325 701 900 1 701 7022 19900 2 14044 70248 585600 561984 23616 12.34= 3.5128 3.513 (v) 8.6666 2. 9 4 3 9 2 8.66 66 00 00 2 4 49 466 9 441 584 2566 3 2336 5883 23000 3 17649 58869 535100 529821 5279 8.6660 = 2.9439 2.944
(vi) 234.234 15. 3 0 4 7 1 2 34.23 40 00 00 1 1 25 134 5 125 303 923 3 909 30604 144000 4 122416 306087 2158400 2142609 15791 234.234 = 15.3047 15.305 3. Find the square root of the following numbers correct to 4 decimal places. (i) 13 3. 6 0 5 5 5 3 13. 00 00 00 00 00 3 9 66 400 6 395 7205 40000 5 36025 72105 397500 5 360525 721105 3697500 3605525 91975 13 = 3.60555 3.605
(ii) 8.12 2. 8 4 9 5 6 2 8. 12 00 00 00 00 2 4 48 412 8 384 564 2800 4 2256 5686 54400 9 51201 56985 319900 5 284925 56990 3497500 569906 3497500 3419436 78064 8.12 = 2.8956 2.8496
(iii) 3333 5 7. 7 3 2 1 4 5 33 33. 00 00 00 00 00 5 25 107 833 7 749 1147 8400 7 8029 11543 37100 3 34629 115462 247100 2 230924 1154641 16171600 1 1154641 11546424 46295900 46185696 110504 3333 = 57.73214 57.7321
4. Find the approximation from below to 4 decimal places to the square root of the following numbers. (i) 5 2. 2 3 6 0 6 7 2 5. 00 00 00 00 00 00 2 1 43 100 2 84 443 1600 3 1329 4466 27100 6 26796 447206 3040000 6 2683236 4472127 35676400 31304889 4371511 5 = 2.236067 (2.236067) 2 = 4.9996<5< 5.000009 = (2.23607) 2 2.236007 is the approximation from below of 5 to 5 decimal places.
(ii) 8 2. 8 2 8 4 2 7 2 8. 00 00 00 00 00 00 2 4 42 100 8 384 562 1600 2 1124 5648 47600 8 45184 56564 241600 4 226256 565682 1534400 2 1131364 5656847 40303600 39597929 705671 8 = 2.828427 (2.82842) 2 = 7.999959 < 8 < 8.000016 = (2.828443) 2 2.82842 is the approximation from below of 8, correct to 5 decimal places.
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 (900+900) 2 = 1800m 2 Side of the new garden is 1800 m. A = side x side = (side) 2 = 42.43m 4 2. 4 2 6 4 18 00.00 00 00 4 16 82 200 2 164 844 3600 4 3376 8482 22400 2 16964 84846 543600 509076 34524
NON-TEXTUAL QUESTIONS 1. Find the square root of the following numbers. (i) 19.5364 19.5364 = 195364 10000 2 195364 2 97682 221 48841 221 19.5364 195364 10000 = (2x221 )2 10000 = 442 100 = 4.42 (ii) 1.993744 1.993744 = 1993744 100000 1993744 1993744 1000000 (4 353 ) 2 1412 = (100 0 )2 = = 1.412 1000
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) 378225 6 1 5 6 378225 6 36 121 182 1 121 1225 6125 6125 0 378225 = 615 (ii) 923521 9 6 1 9 92 35 21 9 81 186 1135 6 1116 1921 1921 1921 0 923521= 961
4. Find the least number to be added to get the perfect square. (i) 456123 6 7 5 6 45 61 23 6 36 127 961 7 889 1345 7223 6725 498 456123 > 675 2 Therefore we find 675 2 675 2 = 456123 456976 456123 853 853 should be added to 456123 to make it a perfect square. (ii) 7891011 2 8 0 9 2 7 891011 4 48x8 389 384 560x0 510 0 5609x9 51011 50481 530
2810 2 = 7896100 7896100 7891011 5089 5089 should be added to 7891011 to make it a perfect square.
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 = 500 22 row and column are possible 22 x 22 = 484 students can perform the drill. Remaining students 500-484 = 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
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 5 2 2 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 = 2 2 2 2 3 2 5 2 = 3600 3600 is a perfect number that is divisible by 8, 15, and 20.
4. Express 13 as sum of two consecutive integer we have n = 13. Ans: n 2 1 2 = 13 2 1 2 = 169 1 2 = 168 2 = 84 n 2 +1 2 = 13 2 + 1 2 = 169 + 1 2 = 168 2 = 85 13 2 = 84+85 169 = 84+85 13 2 can be expressed as sum of 84 and 85
UNIT-1 MULTIPLICTION OF POLYNOMIALS EXERCISE 3.1.2 1. 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
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
(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 2 + 12x + 35
(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 + 3 2 x 7 = 2 + 3 2 x x + 2 x + 3 ( 7 ) = 4 x 2 + 6 x 14 x 21 = 4 x 2 8 x 21
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) 2 +2.2a.3 +b 2 = 4a 2 + 12a + 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 2 + 2.x. 1 x + ( 1 x )2 = x 2 + 2 + 1 x 2
(iv) ( 12a + 6b ) 2 Using (a + b) 2 = a 2 +2ab +b 2 we get a = 12a and b = 6b 12a + 6b ) 2 = ( 12a) 2 + 2. 12 a + 6b + ( 6b) 2 = 12a 2 +2 72 ab +6b 2 = 12a 2 +2 36 2 ab +6b 2 = 12a 2 + 12 2 ab +6b 2 (v) (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 + ( 22 7 )2 = π 2 + 44 π 7 = π 2 + 44π 7 7 + ( 22 7 )2 + 484 49
(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.3 + 3 2 = 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 2 2 + 1 y 2
(ix) ( 10x 5y) 2 Using (a b) 2 = a 2 2ab + b 2 a = 10xand b = 5y) ( 10x 5y) 2 = ( 10x) 2 2. 10x. 5y)+ ( 5y)) 2 = 10x 2 2 50 xy + 5y 2 = 10x 2 2.5 2 xy + 5y 2 = 10x 2 10 2 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 + ( 22 7 )2 = π 2 44 π 7 7 + ( 22 7 )2 = π 2 44π 7 + 484 49
(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) + 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 12 4. 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 2 3 2 = x 2 9
(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 2 9 4 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
(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 2 4 2 ) (x 2 3 2 ) = (x 2 16) (x 2 9) Using (a + b) (a b) = x 2 x (a + b) + ab = (x 2 ) 2 x (16+9) +16.9 = x 4 25x 2 +144
(vii) (x a) (x + a) 1 x 1 a 1 x + 1 a (x 2 a 2 ) 1 + 1 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 2 + 1 1 + x2 a2 a2 x 2 5. Simplify the following: (i) (2x 3y) 2 + 12xy = (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) 2 + 2.3m.5n 4n 2 = 9m 2 + 25n 2 + 30mn 4n 2 = 9m 2 + 30 mn +21n 2
(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 2 + 2. x. 1 x + 1 = x 2 + 2 + 1 = x 2 + 2 + 1 x 2)2 (m 2 + 2. m. 1 m + 1 m 2)2 x 2 (m2 + 2 + 1 m 2) x 2 m2 + 2 1 m 2 = x 2 m 2 + 1 x 2 1 m 2 + 4 (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
(vi) (3a 2) 2 (2a -3) 2 = (9a 2 2.3a.2 + 2 2 ) (4a 2 2.3a.2 + 9 2 ) = 9a 2 12a + 4 4a 2 + 12a 9 = 5a 2 5 =5(a 2 1)
EXERCISE 3.1.3 1. 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 (4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2 = x 3 + 11x 2 + x (20 + 10 + 8) +40 = x 3 + 11x 2 + 38x +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 (3 + 2 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1) = y 3 + 4y 2 + y (6 2 3) 6 = y 3 + 4y 2 + y 6
(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 (2 3 + 4) + a [2( 3) + ( 3)4 + 4.2) + 2 ( 3) 4 = a 3 + 3a 2 + a ( 6 12 + 8) 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 [2 + 6 + 3] 6 = m 3-6m 2 + 11m 6
(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) + 5 + 7 + 2 3. 5 + 5. 7 + 7. 3 + 3. 5. 7 = 2 2 + 2 3) + 5 + 7 + 2 15+ 35 + 21 + 105 (vi) 105 x 101 x 102 We can write this as (100 + 5) (100 + 1) (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 = 1 and c = 2 (100 + 5) (100 + 1) (100 + 2) = 100 3 + 100 2 (5 + 1 + 2) + 100 (5.1 + 1.2 + 2.5) + 5.1.2 = 1000000 + 10000 (8) + 100(5 + 2 +10) + 10 = 1000000 + 80000 + 1700 +10 = 1081710
(vii) 95 x 98 x 103 We can write this as (100-5) (100-2) (100 + 3) 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 + 3) = 100 3 + 100 2 ( 5 2 +3) + 100( 5) ( 2) + ( 2) 3 (viii) 1.01 x 1.02 x 1.03 We can write this as (1 + 0.01) (1 + 0.02) (1 + 0.03) + 3 (-5) + (-5) (-2) 3 = 1000000 + 10000( 4) +100 (10 6 16) + 30 = 1000000 40000 1100 +30 = 958930 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) (1 + 0.02) (1 + 0.03) = 1 3 + 1 2 (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) + (0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03) = 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006 = 1.061106
2. Find the coefficients of x 2 and x in the following: (i) (x + 4) (x + 1) (x + 2) = x 3 + x 2 (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2 = x 3 + 7x 2 + 14x + 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 + 6 + 5) 30 = x 3 12x 2 + 41x 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 [ 2 +10 5] + 10 = 8x 3 24x 2 + 6x + 10 Coefficient of x 2 is 24 and x is 6
(iv) ( x 2 + 1) ( x 2 + 2) ( x 2 + 3) = ( x 2 )3 + ( x 2 )2 [1 + 2 + 3] + x 2 [1.2 + 2.3 + 3.1] + 1.2.3 = x3 8 + x2 4 (6) + x 2 (2 + 6 + 3) + 6 = x3 8 + 3 2 x2 + 11 2 x + 6 Coefficient of x 2 is 3 2 and x is 11 2 3. 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 + 2 3) + 6 = x 3-7x 2 + 6
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 + 2 5] + 10 = x 3 + 2x 2 13x 10 5. 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.
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 3 + 12 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 3 + 36 x 2 y + 54 xy 2 + 27y 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 3 + 240a 2 b + 300ab 2 + 125b 3
(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 2 + 1 x 3 = x 3 + 3x + 3 + 1 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) 3 + 3 (20) 2 (3) + 3(20) 3 2 + 3 3 = 8000 + 3600 + 540 + 27 = 12167 (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) 2 + 1 3
= 125000 + 7500 + 150 +1 = 132651 (vii) 101 3 We write 101 as (100 + 1) 3 Using identity (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 100, b = 1 (100 + 1) 3 = 100 3 + 3. 100 2 + 3. 100.1 2 + 1 3 = 1000000 + 30000 + 300 + 1 = 1030301 (viii) 2.1 3 We write 2.1 (2 + 0.1) 3 Using identity (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 we get a = 2, b = 0.1 (2 + 0.1) 3 = 2 3 + 3 x 2 2 (0.1) + 3 x 2 x (0.1) 2 + (0.1) 3 = 8 + 1.2 + 0.06 + 0.001 = 9.261
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 2 1 + 3x( 1 x x )3 ( 1 x )3 = x 3 3x + 3x x 2 1 x 3 = x 3 3x + 3 1 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) 2.2 + 3. 3x x 2 2 2 3 = 3 3 x 3 6. 3 x 2 + 12 3 x 8 =3 3 x 3 18x 2 + 12 3 x 8
(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) 2 5 + 3. 2x. 5) 2 ( 5) 3 = 8x 3 12 5x 2 + 30x 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 = 1 50 1) 3 = 50 3 3.50 2.1 + 3.50.1 2 1 3 = 125000 3 x 2500 + 150 1 = 125000 7500 +149 = 117649 (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 = 20 3 3.20 2.2 + 3.20.2 2 2 3 = 8000 6x400 + 60x4-8 = 8000 2400 +240 8
= 5832 (vii) 95 3 We write 95 = 100 5 Using (a b) 3 = a 3 3a 2 b + 3ab 2 b 3 we get a = 100, b = 5 (100 5) 3 = 100 3 3.100 2.5 + 3.100.5 2 5 3 = 1000000 150000 + 7500 125 = 857375 (viii) 108 3 We write 108 3 = (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 3 3. (110) 2.2 + 3.110x2 2 2 3 = 1331000 72600 + 1320-8 = 1259712
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 3 + 1 x 3 + 3 (3) = x 3 + 1 x 3 = 27 9 = x 3 + 1 x 3 = 18 9. If p + q = 5 and pq = 6, find p 3 + q 3 (p + q) 3 = p 3 + 3pq (p + q) + q 3 5 3 = p 3 + 3.6 (5) + q 3 125 = p 3 + 90 + q 3 p 3 + q 3 = 125 90 p 3 + q 3 = 35
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 3.10 (3) 27 = a 3 b 3 90 a 3 b 3 = 27 + 90 a 3 b 3 = 117 11. If a 2 + 1 a 2 = 20 and a3 + 1 a 3 = 30, find a + 1 a a 3 + 1 a 3 = (a + 1 a ) (a2 + 1 a 2 - a x 1 a ) 30 = (a + 1 a ) = (20 1) 30 = (a + 1 a ) x 19 30 19 = a + 1 a a + 1 a = 30 19
EXERCISE 3.1.4 1. 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) 2 + 2.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) 2 + 2.p.q + 2q(-2r) +2(-2r)p = p 2 + q 2 + 4r 2 + 2pq 4pr 4pr
(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 2 4 + 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 2 + 9 + 1 m 2-6m + 6 m - 2
= m 2 + 1 m 2 + 6 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 + 5 + 1 2x )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 + 5 + 1 2x )2 = x 2 + 5 2 + ( 1 2x )2 + 2.x.5 + 2.5. 1 2x + 2( 1 2x )x = x 2 + 25 + 1 4x 2 +10x + 5 x + 1 = x 2 + 1 4x 2 +10x + 5 x + 26
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) 2 + 5 2 + 2.3x.4y + 2.4y.5 + 2.5(3x)] [x 2 + (5y) 2 + (-4) 2 + 2. X.5y + 2.5y (-4) + 2 (-4). x] = 9x 2 + 16y 2 + 25 + 24xy + 40y + 30x [x 2 + 25y 2 + 16 + 10xy 40y-8x = 9x 2 + 16y 2 + 25 + 24xy + 40y + 30x - x 2-25y 2-16 - 10xy + 40y + 8x = 8x 2 9y 2 14xy + 80y + 38x + 9
(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) 2 + 2.2m( 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 2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y) 2 + (3z) 2 (r) 2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y) = 2(x + 2y) 2 + 9z 2 + r 2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z 2 + r 2 + 6 (x + 2y) z - 6zr 2r (x + 2y) = 2(x 2 + 2.x.2y +4y 2 ) + 18z 2 + 2r 2 + 12 (x + 2y)z = 2x 2 + 8xy + 8y 2 + 18z 2 + 2r 2 +12xz + 24 yz = 2x 2 + 8y 2 + 18z 2 + 2r 2 + 8xy +12xy + 24 yz
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 = 50 50 + 2ab + 2bc + 2ca = 144 2(ab + bc + ca) = 144 50 2(ab + bc + ca) = 94 ab + bc + ca = 94 2 ab + bc + ca = 47 4. 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) = 35 + 2 (23) = 35 + 46 (a + b + c) 2 = 81 (a + b +c) = ± 81 a + b +c = ± 9
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 = 5 21 + 3y = 5 15x 3y = 12 2 + 3y = 5 17x = 17 3y = 5 2 x = 17 17 3y = 3 x = 1 y = 3 3 = 1 x = 1 y = 1
EXERCISE 3.1.5 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
(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
(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
(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
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
(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)
= 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)
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.
(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.
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.
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)
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
8. If x 2 3x + 1 = 0, prove that x 2 + 1 = 7. x2 Given x 2 3x + 1 = 0 x 2 + 1 = 3x x + 1 x = 3 (dividing both sides by x) Squaring both sides we get (x + 1 x )2 = 3 2 = x 2 + 1 x 2 + 2x. 1 x = 9 x 2 + 1 x 2 + 2 = 9 x 2 + 1 x 2 = 9 2 = 7
ADDITIONAL PROBLEMS ON MULTIPLICATION OF POLYNOMIALS I. Find the following products: (i) (2a + 3b) (4a 2 6ab + 9b 2 ) = 8a 3 + 12a 2 b 12ab 2 18ab 2 + 18ab 2 + 27b 3 = 8a 3 + 27b 3 (ii) (3x + 4y) (9x 2 12xy +16y 2 ) = 27x 3 + 36x 2 y - 36x 2 y - 48xy 2 + 48xy 2 + 64y 3 = 27x 3 + 64y 3 (iii) (5x 2y) (25x 2 + 10xy + 4y 2 ) = 125x 3 50x 2 y + 50x 2 y 20xy 2 + 20xy 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
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) 3 + 1 = (2x + 1) [(2x) 2 2x.1 + 1 2 ] 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) 3 + 3 3 = (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) 3 + 5 = (3a + 5) (9a 15a + 25) Hence (3a + 5) should be multiplied.
(iv) 4a + 3 We have a 3 + b 3 = (a + b) (a 2 ab +b 2 ) a = 4a b = 3 (4a) 3 + 3 = (4a + 3) [(4a) 2 4a.3 + 3 2 ) = (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) 3 3 3 = (5a 3 ) [(5a 2 ) + (5a) 3 + 3 2 ] 125a 3 27 = (5a 3) (25a 2 + 15a + 9) (25a 2 + 15a + 9) should be multiplied. (ii) `16a 2 + 20a + 25 We have a 3 - b 3 = (a - b) (a 2 + ab +b 2 ) a = 4a b = 5 (4a) 3 5 3 = (4a 5 ) (4a 2 + 20a + 25) (4a 5) should be multiplied.
(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) 3 5 3 = (4x 5 ) (4x 2 + 20x + 25) (4x 5) should be multiplied. 4. Use the appropriate identity to compute the following: (i) (103) 2 We write 103 = 100 + 3 Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 100 b = 3 (100 + 3) 2 = 100 2 + 2.100.3 +3 2 = 10000 + 600 + 9 = 10609
(ii) (107) 2 We write 107 = 100 + 7 Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 100 b = 7 (100 + 7) 2 = 100 2 + 2.100.7 +7 2 = 10000 + 1400 + 49 = 11449 (iii) ( 50 1 2 )2 We write 50 1 2 = 50.5 = 50. +0.5 Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 50 b = 0.5 (50 + 0.5) 2 = 50 2 + 2 x 50(0.5) + (0.5) 2 = 2500 + 100 (0.5) +0.25 = 2500 + 50 +0.25 = 2550.25 (iv) (998) 2 We write 998 = 1000-2 Using (a + b) 2 = a 2 + 2ab + b 2 We get a = 1000 b = -2 (1000-2) 2 = 1000 2-2.1000.2 +2 2
= 1000000 4000 + 4 = 996004 (v) 107 x 93 We write 107 = 100 + 7 and 93 = 100-7 Using (a + b) (a b) = a 2 + b 2 We get a = 100 b = 7 (100 + 7) (100 7) = 100 2 7 2 = 10000 49 = 9951 (vi) 1008 x 992 We write 1008 = 1008 + 8 and 992 = 1000-8 Using (a + b) (a b) = a 2 + b 2 We get a = 1000 b = 8 (1000 + 8) (1000 8) = 1000 2 8 2 = 100000 64 = 99936
(vii) (101) 3 We write 101 = 100 + 1 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 We get a = 100 b = 1 (100 + 1) 3 = 100 3 + 3(100) 2.1 + 3(100).1 2 + 1 3 = 1000000 + 30000 + 300 +1 = 10303001 (viii) (1002) 3 We write 1002 = 1000 + 2 Using (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 We get a = 1000 b = 2 (1000 + 2) 3 = 1000 3 + 3(1000) 2.2 + 3(1000).2 2 + 2 3 = 1000000000 + 6000000 + 12000 + 8 = 1006012008
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 2 + 2.12 = 49 x 2 + y 2 = 49 24 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 = 343 252 x 3 + y 3 = 91
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 2 + 1 x 2 + 2.x. 1 x = 9 x 2 + 1 x 2 = 9 2 x 2 + 1 x 2 = 7 Squaring again (x 2 + 1 x 2 )2 = 7 2 x 4 + 1 x 4 + 2.x2. 1 x 2 = 49 x 4 + 1 x 4 = 49 2 x 4 + 1 = 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
a 3 b 3 = 8 + 90 a 3 b 3 = 98 8. 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)
= (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 2 + 2 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
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 2 + 6-1 = 184. x 2 x3 Given x 1 x = 4 Squaring both sides (x 1 x )2 = 4 2 x 2 + 1 x 2-2x. 1 x = 16 x 2 + 1 x 2 = 16 + 2 x 2 + 1 x 2 = 18..(1)
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 = 64 + 12 x3 x 3 1 x 3 = 76..(2) Now consider L.H.S x 3 + 6x 2 + 6 x 2-1 x 3 = x 3 1 + 6 x 3 (x2 + 1 ) x 2 = 76 + 6 x 18 = 76 + 108 = 184 R.H.S
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]
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
15. Suppose a, b, are the sides of a triangle such that 2s = a + b + c. Prove that 1 s a + 1 + 1 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
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) 2 + 2.3a.5b + (5b) 2 = 9a 2 + 30ab + 25b 2 (ii) ( 1 2 x + 2 3 y)2 Using (a + b) 2 = a 2 + 2ab + b 2 we get a = 1 2 x b = 2 3 y ( 1 2 x + 2 3 y)2 = ( 1 2 x)2 + 2. x 2. 2y 3 + ( 2y 3 )2 (iii) (2a + 3b + 4c) 2 = x2 + 2 4y2 xy + 4 3 3 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) 2 + 2.2a.3b + 2.3b.4c + 2.4c2a = 4a 2 + 9b 2 + 16c 2 + 12ab + 24bc + 16ca
(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 2 + 25c 2 12ab 20bc + 30c II. If (x 1 ) = 3, find the valve of 2x (i) x 2 + 1 4x 2 (ii) x4 + 1 16x 4 (iii) x3 1 8x 3 (i) (x 1 2x ) = 3 Squaring on both sides (x 1 2x )2 = 3 2 x 2 + 1 4x 2 2.x. 1 2x = 9 x 2 + 1 4x 2 = 9 + 1 = 10
(ii) x 2 + 1 4x 2 = 10 Squaring on both sides (x 2 + 1 4x 2 )2 = 10 2 x 4 + x 4 + 1 + 16 x 4 2.x2 x 1 = 100 4x 2 1 16 x 4 + 1 2 = 100 1 x 4 + = 100-1 = 99 1 16 x 4 2 2 (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 = 27 + 9 = 54 + 9 = 63 8x 3 2 2 2
3. If a 2 + 1 = 14, find the values of a2 (i) a + 1 a (ii) a 1 a (iii) a 2 1 a 2 (i) a 2 + 1 a 2 = 14 Additing 2 on both sides (a 2 + 1 + 2) = 14 + 2 a2 (a 2 + 1 a 2)2 = 16 a + 1 a = 12 = ±4 (ii) a 2 + 1 a 2 = 14 Subtracting 2 on both sides (a 2 + 1 2) = 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)
4. If (3a + 4b) = 16 and ab = 4 find the value of 9a 2 + 16b 2 We have (3a + 4b) = 16 Squaring on both sides (3a + 4b) 2 = 16 2 9a 2 + 2.3a.4b + 16b 2 =256 9a 2 + 16b 2 = 256 24ab = 256 24 x 4 = 256-96 = 160 5. 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 = 1 + 4 (a + 2) 2 = 5 a + 2 = 5 1 a = 1 5 2 a = 5 2
(i) a + 1 a = 5 2 + 1 5 2 = ( 5 2)2 + 1 5 2 = ( 5 2)2 + 1 5 2 = 5 + 4 2.2 5 + 1 5 2 = 10 4 5 5 2 (ii) a 1 a = 5 2-1 5 2 = ( 5 2)2 1 5 2 = 5 + 4 2 5 2 1 5 2 = 9 4 5 1 5 2 = 9 4 5 5 2 = 4(2 5 ) (2 5 ) = 4
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 3 + 27c 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 2 + 1 + k) (k 2 +1 - 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)
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 + 32) (968 b) = 1000 x 936 = 936000 (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 = (98.7 + 1.3) (98.7 1.3) = 100 x 97.4 = 9740
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 2 + 2 x 12 = 64 x 2 + y 2 = 64 24 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 4 + 2 x 144 = 1600 x 4 + y 4 = 1600 288 x 4 + y 4 = 1312 10. If a + b = 8 and ab = 15, find a 3 + b 3. Take a + b = 8 Cubing both sides
(a + b) 3 = 8 3 a 3 + b 3 + 3ab (a + b) = 512 a 3 + b 3 + 3 x 15(8) = 512 a 3 + b 3 + 360 = 512 a 3 + b 3 = 512 360 a 3 + b 3 = 152 11. 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 + 1 + 3(3) = 27 x3 x 3 + 1 x 3 = 27 9 x 3 + 1 x 3 = 18 12. 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
x 3 ( 1 x )3 3x + 1 x (x 1 x ) = 125 x 3 1 3(5) = 125 x3 x 3 1 15 = 125 x3 x 3 1 = 125 + 15 x3 x 3 1 x 3 = 140 13. 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
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 2 + 1 x 2 + 2 x 1 x =9 x 2 + 1 x 2 = 9 2 x 2 + 1 x 2 = 7..(1) Now (x 2 1 x )2 = (x 1 x )2 4x x 1 x (x 1 x )2 = 3 2 4 (x 1 x )2 = 5 x 1 = 5 (2) x
Now x 2 12 x = (x + 1 x ) (x 1 x ) From (1) and (2) x 2 + 1 = 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
UNIT-1 POLYGONS EXERCISE 4.1.3 1. 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
(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