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UNIT-1 SQUARE ROOT EXERCISE 1.1.1 1. Find the square root of the following numbers by the factorization method (i) 82944 210 x 34 = (25)2 x (32)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 = (25 )2 × (32 )2 = 25 x 32 = 288 (ii) 155236 2 155236 2 77618 197 38809 197 197 1 = 22 X 1972 = 155236 = (2 × 197)2 = 2 X 197 = 394 (iii) 19881 3 19881 3 6627 47 2209 47 = 19881 = 32 × 472 = 3 x 47 = 141 2. Find the square root of the following numbers. (i) 184.96 184 .96 x 100 100 = 18496 100 = (2 3 )2 × 17 2 2 18496 2 9248 2 4624 2 2312 2 1156 2 578 289 17 289 17 = = = β΄ 184.96 (2 3 )2 × 17 2 10 2 8x17 10 136 10 = 13.6 10 2 (ii) 19.5364 19.5364 = 2 195364 2 97682 221 48841 221 19.5364 = = = 195364 10000 ( 2 ×221 )2 100 2 442 100 = 4.42 195364 10000 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? 92 = 81 digits-2 102 = 100 digits-3 992 = 9801 digits-4 1002 = 10000 digits-5 9902 = 998001 digits-6 10002 = 1000000 digits-7 99992 = 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 π 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 β΄ 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 432 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 > 782 Therefore we find 792 792 = 6241 6241 6200 41 (ii) 12675 112 1 1 26 75 1 1 21 26 1 21 222 575 444 131 β΄ 41 is the least number to be added to get a perfect square. 1132 = 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 2982 = 88804 88804 88417 387 β΄387 is the least number to be added to 88417 to make it a perfect square. (iv) 123456 351 3 12 34 56 3 9 65 334 5 325 701 956 701 255 123456 lies between 3512 and 3522 3512 = 123201 3522 = 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 35 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 65 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 185 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 331 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 67 6 45 67 6 36 127 967 889 78 The number 4567 lies between 672 and 682. 672 = 4489 682 = 4624 (ii) 56789 238 2 5 67 89 2 4 43 167 3 129 468 3889 3744 145 56644 lies between the squares of 2382 and 2392. 2382 = 56644 2392 = 57121 (iii) 88888 298 2 8 88 88 2 4 498 488 441 4788 3984 804 88888 lies between the squares of 2982 and 2992. 2982 = 88804 2992 = 89401 (iv) 123456 351 3 12 34 56 3 9 65 334 5 325 701 956 701 255 123456 lies between 3512 and 3522 3512 = 123201 3522 = 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. 154 1 2 39 40 1 1 25 139 5 125 304 1440 1216 224 The area of the new square plot is 1542 = 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 (ii) (1.234)2 (1.234)2 = 1.5227556 (iii) Ans: 6 (0.0023)2 (0.0023)2 = 0.00000529 (iv) Ans: 4 Ans: 8 (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 0.085849 = 0.293 (iii) Ans: 3 1.844164 1.844164= 1.358 (iv) Ans: 4 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 24 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 00 06 043 6 36 125 716 5 625 1307 9149 9149 0 1 (iii) 0.431649= 0.657 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 24686.6944m2. 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 24686.6944m2 Length of a side of the garden the is 24686.6644in 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) = 24062.2144m2 EXERCISE 1.1.4 1. Round off the following numbers to 3 decimal places. 1. 1.5678 β 1.568 2. 2.84671 β 2.847 3. 14.56789 β 14.568 4. 12.987564 β 12.988 5. 3.3333567 β 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 39 64 300 4 256 686 4400 6 4115 6924 28400 4 27696 69281 80400 69281 11119 12 = 3.4641 β 3.464 (ii) 1.8 1.3 4 1 6 1 1.80 00 00 00 11 23 080 3 69 264 1100 4 1056 2681 4400 1 2681 26826 171900 160956 10644 1.8 = 1.3416 β 1.3416 = 1.342 (iii) 133 1 1. 5 3 2 5 1 1 33. 00 00 00 00 11 21 033 1 21 225 1200 5 1125 2303 7500 3 6909 23062 59100 2 46124 230645 1297600 1153225 144375 133 = 11.5325 β 11.533 (iv) 12.34 3. 5 1 2 8 3 12.34 00 00 00 39 65 334 5 325 701 900 1 701 7022 19900 2 14044 70248 585600 561984 23616 12.34= 3.5128 β 3.513 (v) 8.6666 2. 9 4 3 9 2 8.66 66 00 00 24 49 466 9 441 584 2566 3 2336 5883 23000 3 17649 58869 535100 529821 5279 8.6660 = 2.9439 β 2.944 (vi) 234.234 15 . 3 0 4 7 1 2 34.23 40 00 00 11 25 134 5 125 303 923 3 909 30604 144000 4 122416 306087 2158400 2142609 15791 234.234 = 15.3047 β 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 β 3.605 (ii) 8.12 2 2 48 8 564 4 5686 9 56985 5 56990 569906 2. 8 4 9 5 6 8. 12 00 00 00 00 4 412 384 2800 2256 54400 51201 319900 284925 3497500 3497500 3419436 78064 8.12 = 2.8956 β 2.8496 (iii) 3333 5 5 107 7 1147 7 11543 3 115462 2 1154641 1 11546424 5 7. 7 3 2 1 4 33 33. 00 00 00 00 00 25 833 749 8400 8029 37100 34629 247100 230924 16171600 1154641 46295900 46185696 110504 3333 = 57.73214 β 57.7321 4. Find the approximation from below to 4 decimal places to the square root of the following numbers. (i) 5 2 2 43 2 443 3 4466 6 447206 6 4472127 2. 2 3 6 0 6 7 5. 00 00 00 00 00 00 1 100 84 1600 1329 27100 26796 3040000 2683236 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 2 42 8 562 2 5648 8 56564 4 565682 2 5656847 2. 8 2 8 4 2 7 8. 00 00 00 00 00 00 4 100 384 1600 1124 47600 45184 241600 226256 1534400 1131364 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 900M2. 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 900m2. If additional land, measuring equal area to added to the garden, its area will be (900+900)2 = 1800m2 Side of the new garden is 1800 m. A = side x side = (side) 2 = 42.43m 4 4 82 2 844 4 8482 2 84846 4 2. 4 2 6 18 00.00 00 00 16 200 164 3600 3376 22400 16964 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 (ii) (2x221 )2 = 10000 = 442 100 = 4.42 1.993744 1.993744 = 1993744 100000 1993744 1993744 1000000 = (4 × 353 )2 (100 0 )2 = 1412 1000 = 1.412 2. If a perfect square A has n digits, how many digits do you expect in π? 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 > 6752 Therefore we find 6752 6752 = 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 28102 = 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 β 500 x2 = 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 = x2 x2 = 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 22 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 = 22 22 32 52 = 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 β1 2 n 2 +1 2 = = 13 2 β 1 2 13 2 + 1 2 = = 169 β 1 2 169 + 1 2 = = 168 2 168 2 = 84 = 85 132 = 84+85 169 = 84+85 132 can be expressed as sum of 84 and 85 UNIT-1 MULTIPLICTION OF POLYNOMIALS EXERCISE 3.1.2 1. Evaluate the following products: (i) ax2 ( bx + c) = ax2 (bx) + ax2 (c) = abx3 + acx2 (ii) ab (a+b) = ab (a) + ab (b) = a2b + ab2 (iii) a2b2 (ab2+a2b) = a2b2 (ab2) + a2b2 (a2b) = a3b4 + a4b3 (iv) b4(b6 + b8) = b4 (b6) + b4 (b8) = b10 + b12 2. Evaluate the following products: (i) (x+3) (x+2) = (x+3) x + (x+2) x = x2 + 3x + 2x + 6 = x2 + 5x + 6 (ii) (x+5) (xβ2) = (x+5) x + (x+5) (β2) = x2 +5x β 2x β10 = x2 + 3x β10 (iii) ( yβ4 ) ( y+6 ) = (y β 4) y +(y β 4)6 = y2 β 4y + 6y β 24 = y2 + 2y β 24 (iv) (aβ5) (aβ6) = (aβ5) a + (aβ6) (β6) = a2 β 5a β 6a + 30 = a2 β 11a +30 (v) (2x+1) (2xβ3) = (2x+1)2x + (2x+1) (β3) = 4x2 + 2x β 6x β 3x = 4x2 β 4x β3 (vi) (a +b)(c +d) = (a + b) c + (a+b) d = ac + bc + ad + bd (vii) ( 2x β 3y ) ( x β y ) = (2x β 3y) x + (2x β 3y) (βy) = 2x2 β 3xy β 2xy +3y2 = 2x2 β 5xy + 3y2 (viii) ( ππ± + π ) ( ππ± + π ) = ( ππ± + π ) ( ππ±) + ( ππ± + π ) ( π) = ππ x2 + 5x + 7x + ππ = ππx2 + 12x + ππ (xi) (2a+3b) (2aβ3b) = (2a+3b) 2a + (2a+3b) (β3b) = 4a2+ 6ab β 6ab β 9b2 = 4a2 β 9b2 (xii) (6xyβ5) (6xy+5) = (6xy β 5) (6xy) + (6xy β 5) 5 = 36x2y2 β 30xy + 30xy β 25 = 36x2y2 β 25 2 (xiii) x 2 = = = 2 +3 x 4 x2 4 x2 x 2 +3 x 6 14 x x + β 8 β β21 x β7 + β 21 2 x + 3 (β 7 ) 3. Expand the following using appropriate identity: (i) (a +5)2 Using (a + b)2 = a2 +2ab +b 2 we get a=a (a +5)2 b=5 = a2 +2.a.5 +b 2 = a2 +10a +25 (ii) (2a +3)2 Using (a + b)2 = a2 +2ab +b 2 we get a = 2a b=3 (2a +3)2 = (2a)2 +2.2a.3 +b 2 = 4a2 + 12a + 9 (iii) π ( x + )2 π± Using (a + b)2 = a2 +2ab +b 2 we get a = 2a b= 1 x 1 1 1 x x x (X + )2 = x2 + 2.x. + ( = x2 + 2 + π π±π )2 (iv) ( 12a + 6b )2 Using (a + b)2 = a2 +2ab +b 2 we get a = 12a and b = 6b 12a + 6b )2= ( 12a)2 + 2. 12 a + 6b + ( 6b)2 = 12a2 +2 72 ab +6b 2 = 12a2 +2 36 × 2 ab +6b 2 = 12a 2 + 12 π ab +6b2 (v) (π + ππ 2 ) π Using (a + b)2 = a2 +2ab +b 2 we get a = π and b = (π + 22 2 ) 7 = π2 + 2. π = π2 + 44 Ο = π2 + πππ 7 π 22 7 22 7 22 + ( )2 7 22 + ( )2 7 + πππ ππ (vi) (y β 3)2 Using (a β b)2 = a2 β 2ab + b 2 a = y and b = β3 (y β 3)2 = y2 β 2. y.3 + 32 = y2 β 6y + 9 (vii) (3a β 2b)2 Using (a β b)2 = a2 β 2ab + b 2 a = 3a and b = β2b (3a β 2b)2 = (3a)2 β 2.3a.2b + (2b)2 = 9a2 β 12ab + 4b2 π (viii) ( y β )2 π² Using (a β b)2 = a2 β 2ab + b 2 a = y and b = 1 y 1 1 y y (y β )2 = y2 β 2.y. + ( = y2 β 2 + π π²π 1 2 ) y (ix) ( 10x β 5y)2 Using (a β b)2 = a2 β 2ab + b 2 a = 10xand b = 5y) ( 10x β 5y)2 = ( 10x)2 β 2. 10x. 5y)+ ( 5y))2 = 10x2 β 2 50 xy + 5y2 = 10x2 β 2.5 2 xy + 5y2 = 10x2 β 10 2 xy + 5y2 (x) (π β ππ 2 ) π Using (a + b)2 = a2 +2ab +b 2 we get a = π and b = (π β 22 2 ) 7 22 7 = π2 β 2. π 22 = π2 β 44 Ο 22 = π2 β πππ 7 π 7 +( + ( )2 7 + πππ ππ 22 7 )2 (xi) (2x+3) (2x+5) Using (x + a) (x + b) = x2 + x (a + b) ab we get x = 2x, a = 3 and b = 5 (2x+3) (2x+5) = (2x)2 +2x (3 + 5) + 3.5 = 4x2 +16x +15 (xii) (3x β 3) (3x + 4) Using (x + a) (x + b) = x2 + x (a + b) ab we get x = 3x, a = β3 and b = 4 (3x β 3) (3x + 4) = (3x)2 + 3x [(β3)+(4)] + (-3)4 = 9x2 + 3x β12 = 9x2 + 3x β12 4. Expand : (i) (x + 3 ) (x β 3) Using (a + b) (a β b) = a2 β b2 we get a = x, b = 3 (x + 3) (x β 3) = x2 β 32 = x2 β 9 (3x β 5y) (3x + 5y) (ii) Using (a + b) (a β b) = a2 β b2 we get a = 3x, b = 5y (3x β 5y) (3x + 5y) = (3x)2 β (5y)2 = 9x2 β 25y2 (iii) x 3 + y x 2 3 + y 2 Using (a + b) (a β b) = a2 β b2 we get x y 3 2 a= ,b= x 3 + y x 2 3 + y 2 = (iv) =( π±π π x 2 ) 3 + -( y 2 )2 π²π π (x2 + y2) (x2 β y2) Using (a + b) (a β b) = a2 β b2 we get a = x2, b = y2 (x2 + y2) (x2 β y2) = (x2)2 β (y2)2 = x4 β y4 (v) (a2 + 4b2) (a + 2b) (a - 2b) Using (a + b) (a β b) = a2 β b2 for 2nd and 3rd term we get (a2 + 4b2) (a + 2b) (a - 2b) = (a2 + 4b2) [a2 β (2b)2] = (a2 + 4b2) (a2 β 4b2) Using the above identity once again we get = (a2)2 β (4b2)2 = a4 β 16b4 (vi) (x β 4) (x + 4) (x β 3) (x + 4) Using (a + b) (a β b) = a2 β b2 we get (x β 4) (x + 4) (x β 3) (x + 4) = (x2 β 42) (x2 β 32) = (x2 β 16) (x2 β 9) Using (a + b) (a β b) = x2 β x (a + b) + ab = (x2)2 β x (16+9) +16.9 = x4 β 25x2 +144 π (vii) (x β a) (x + a) 1 (x2 β a2) x2 x 1β 1+ 1 x2 x2 a2 π±π ππ x2 + β x2 x β β a2 x2 π± β π π π π± + π π 1 a2 1 a2 β a2 x 1 x2 + a2 x +1 ππ π±π 5. Simplify the following: (i) (2x β 3y)2 + 12xy = (2x)2 + (3y)2 β 2.2x.3y + 12xy = 4x2 + 9y2 -12xy +12xy = 4x2 + 9y2 (ii) (3m + 5n)2 β (2n)2 = (3m)2 + (5n)2 + 2.3m.5n β 4n2 = 9m2 + 25n2 + 30mn β 4n2 = 9m2 + 30 mn +21n2 1 a2 (iii) (4a β 7b)2 β (3a)2 = (4a)2 β 2.4a.7b + (7b)2 β (3a)2 = 16a2 β 56ab + 49b 2 - 9a2 = 7a2 -56ab + 49b2 (iv) (x + π 2 ) π± (m + π π¦ 1 = (x2 + 2. x. + x = x2 + 2 + = x2 + 2 + 1 x2 1 x2 = x2 β m2+ (v) )2 1 2 ) x2 β (m2 + 2. m. β (m2 + 2 + β m2 + 2 β π π±π β π π¦π 1 m2 1 m2 +4 (m2 + 2n2)2 β 4m2n2 = m4 +2m4.2n2 +4n4 β 4m2 n2 = m4+ 4m2n2 + 4n2 β 4m2n2 = m4 β 4n2 ) 1 m + 1 2 ) m2 (vi) (3a β 2)2 β (2a -3)2 = (9a2 β 2.3a.2 + 22) β (4a2 β 2.3a.2 + 92) = 9a2 β12a + 4 β 4a2 + 12a β 9 = 5a2 β 5 =5(a2 β 1) EXERCISE 3.1.3 1. Find the following products: (i) (x + 4) (x + 5) (x + 2) Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get a = 4, b = 5 and c =2 (x + 4) (x + 5) (x + 2) = x3 + x2(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2 = x3 + 11x2 + x (20 + 10 + 8) +40 = x3 + 11x2 + 38x +40 (ii) (y + 3) (y + 2) ( y β 1) Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get x = y, a = 3, b = 2 and c = -1 (y + 3) (y + 2) (y β 1) = y3 + y2 (3 + 2 β 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1) = y3 + 4y2 + y (6 β 2 β 3) β 6 = y3 + 4y2 + y β 6 (iii) (a + 2) (a β 3) (a + 4) Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get x = a, a = 2, b = β3 and c = 4 (a + 2) (a β 3) (a β 4) = a3 + a2 (2 β 3 + 4) + a [2(β3) + (β3)4 + 4.2) + 2 (β3) 4 = a3 + 3a2 + a (β6 β 12 + 8) β 24 = a3 + 3a2 β 10a β 24 (iv) (m β 1) (m β 2) (m β 3) Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get x = m, a = β1, b = β2 and c = β3 (m β 1) (m β 2) (m β 3) = m3 + m2 (β1 β 2 β 3) + m [(β1) (β2) + (β2) (β3) + (β 3)(β1)] + (β1) (β2) (β3) = m3 + m2 (β6) + m [2 + 6 + 3] β 6 = m3 - 6m2 + 11m β 6 (v) ( π + π) ( π+ π) ( π+ π ) Using (x + a) (x + b) (x + c) = x3 + x2 (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 + =2 2+2 3) + 7 + 2 5. 7 + 7. 3 + 3. 5. 7 3) + 5 + 15 + 35 + (vi) 5+ 7 + 2 21 + 105 105 x 101 x 102 We can write this as (100 + 5) (100 + 1) (100 + 2) Using (x + a) (x + b) (x + c) = x3 + x2 (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) = 1003 + 1002 (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) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get x = 100, a = -5, b = -2 and c = 3 (100 β 5) (100 β 2) (100 + 3) = 1003 + 1002 (β5 β 2 +3) + 100(β5) (β2) + (β2) 3 + 3 (-5) + (-5) (-2) 3 = 1000000 + 10000(β4) +100 (10 β 6 β16) + 30 = 1000000 β 40000 β 1100 +30 = 958930 (viii) 1.01 x 1.02 x 1.03 We can write this as (1 + 0.01) (1 + 0.02) (1 + 0.03) Using (x + a) (x + b) (x + c) = x3 + x2 (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) = 13 + 12 (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 x2 and x in the following: (i) (x + 4) (x + 1) (x + 2) = x3 + x2 (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2 = x3 + 7x2 + 14x + 8 Coefficient of x2 is 7 Coefficient of x is 14 (ii) (x β 5) (x β 6) (x β 1) = x3 + x2 (β5 β 6 β1) + x [(β5) (β6) + (β6) (β1) β 1(β5)] + (β5) (β6) (β1) = x3β 12x2 + x (30 + 6 + 5) β 30 = x3β 12x2 + 41x β 30 Coefficient of x2 is β12 and x is 41 (iii) (2x + 1) (2x β 2) (2x β 5) = (2x)3 + (2x)2 [1β2β5] + 2x [(1)(β2) + (β2)(β5) + (β5)(1)] + 1 (β2) (β5) = 8x3 + 4x2 (β6) + 2x [β2 +10β5] + 10 = 8x3 β 24x2 + 6x + 10 Coefficient of x2 is β24 and x is 6 π± π± π± π π π (iv) ( + 1) ( + 2) ( + 3) x x x 2 2 2 = ( )3 + ( )2 [1 + 2 + 3] + = = x3 8 x3 8 + x2 4 [1.2 + 2.3 + 3.1] + 1.2.3 x (6) + (2 + 6 + 3) + 6 2 3 11 2 2 + x2 + x+6 π ππ π π Coefficient of x2 is and x is 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 β 2) (x β 1) Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get V = x3 + x2 (3 β 2 β 1) + x [3(β2) + (β2) (β1) + (β1)3] + 3(β2) (β1) = x3 - 0x2 + X (β6 + 2 β 3) + 6 = x3 - 7x2 + 6 4. The length, breadth and height of a metal box are cuboid are (x +5), (x β 2) and (x β 1) respectively. What is its volume? Volume of the metal box = length x breadth x height V = (x +5) (x β 2) (x β 1) Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get V = x3 + x2 + (β5 β 2 β1) + x [5(β2) + (β2) (β1) + (β1)5] + 5 (β2) (β1) = x3 + 2x2 + x [β10 + 2 β 5] + 10 = x3 + 2x2 β 13x 10 5. Prove that (a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) β abc [Hint: write a + b = a + b + c β c, b + c = a + b + c β a, c + a = a + b + c β d] x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get L. H.S. = (a + b + c)3 + (a + b + c)2 (βc β a β b) + (a + b + c) [(βc) (βa) + (βa) (βb) + (βb) (βc)] β (βc) (βa) (βb) = (a + b + c)3-(a + b + c)2[(a + b + c)] +(a + b + c) (ac + ab + bc) β abc = (a + b + c)3- (a + b + c)3 + (a + b + c) (ac + ab + bc) β abc = (a + b + c) (ac + ab + bc) β abc = R. H. S. 6. Find the cubes of the following: (i) (2x +y)3 Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 2x, b = y (2x +y)3 = (2x)3 + 3(2x)2 y + 3 (2x) y2 +y3 = 8x3 + 12 x2y + 6 xy2 +y3 (ii) (2x + 3y)3 Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 2x, b = 3y (2x + 3y)3 = (2x)3 + 3(2x)2 (3y) + 3 (2x) (3y)2 + (3y)3 = 8x3 + 36 x2y + 54 xy2 + 27y3 (iii) (4a + 5b)3 Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 4a, b = 5b (4a + 5b)3 = (4a)3 + 3(4a)2(5b) + 3(4a)(5b)2 + (5b)3 = 64a3 + 240a2b + 300ab 2 + 125b 3 π± (iv) ( x + )3 π Using (a + b)3 = a3 + 3a2 b + 3ab2 + b3 we get a = x, b = x 1 x x 1 1 (x + )3 = x3 + 3x2 = x3 + 3x + = x3 + 3x + (v) x x 1 1 + 3x ( )3 + ( )3 3x 1 x x3 3 x 2 + + 1 x3 233 We write this as (20 + 3)3 Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 20, b = 3 (20 + 3)3 = (20)3 + 3 (20)2(3) + 3(20) 32 + 33 = 8000 + 3600 + 540 + 27 = 12167 (vi) 513 We write this as (50 + 1)3 Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 50, b = 1 (50 + 1)3 = (50)3 β 3 x (50)2 x 1 + 3(50) (1)2 + 13 = 125000 + 7500 + 150 +1 = 132651 (vii) 1013 We write 101 as (100 + 1)3 Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 100, b = 1 (100 + 1)3 = 1003 + 3. 1002 + 3. 100.12 + 13 = 1000000 + 30000 + 300 + 1 = 1030301 (viii) 2.13 We write 2.1 (2 + 0.1)3 Using identity (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a = 2, b = 0.1 (2 + 0.1)3 = 23 + 3 x 22(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 β 3b)3 Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = 2a, b = 3b (2a β 3b)3 = (2a)3 β 3 (2a)2(3b) + 3 (2a)(3b)2 β (3b)3 = 8a3 β 36a2b + 54ab2 -27b (ii) π ( x β )3 π± Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = x, b = 1 x 1 1 x x (x β )3 = x3 β 3x2 (iii) 1 1 x x + 3x( )3 β ( )3 = x3 β 3x + 3x = x3 β 3x + 3 x2 x β β 1 x3 1 x3 (β3 x β 2)2 Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = β3 x, b = 2 (β3 x β 2)2 = (β3x)3 β 3 (β3x)2 .2 + 3. β3x x 22 β 23 = 3β3 x3 β 6. 3 x2 + 12β3 x β 8 =3β3 x3 β 18x2 + 12β3 x β 8 (iv) (2x β β5)3 Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = 2x, b = β5 (2x β β5)3 = (2x)3 β 3(2x)2 β5 + 3. 2x. β5)2 β (β5)3 = 8x3 β 12β5x2 + 30x β 5β5 (v) 493 We can write 49 = 50 β 1 Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = 50, b = 1 50 β 1)3 = 503 β 3.502.1 + 3.50.12 β 13 = 125000 β 3 x 2500 + 150 β 1 = 125000 β 7500 +149 = 117649 (vi) 183 We wrote 18 = 20 β 2 Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = 20, b = 2 (20 β 2)3 = 203 β 3.202.2 + 3.20.22 β 23 = 8000 β 6x400 + 60x4 - 8 = 8000 β 2400 +240 β 8 = 5832 (vii) 953 We write 95 = 100 β 5 Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = 100, b = 5 (100 β 5)3 = 1003 β 3.1002.5 + 3.100.52 β 53 = 1000000 β 150000 + 7500 β 125 = 857375 (viii) 1083 We write 1083 = (110 β 2) Using (a β b)3 = a3 β 3a2b + 3ab 2 β b3 we get a = 110, b = -2 (110 β 2)3 = 1103 β 3. (110)2.2 + 3.110x22 β 23 = 1331000 β 72600 + 1320 -8 = 1259712 π π π± π±π 8. If x + = 3, prove that x3 + = 18. 1 Given x + = 3 x Cubing both sides we get 1 (x + )3 = 33 x Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 we get a=x b= 1 x 1 1 1 1 x x x x (x + )3 = (x)3 + ( )3 + 3x. (x + ) = 27 = x3 + = x3 + = x3 + 1 x3 1 x3 1 x3 + 3 (3) = 27 β 9 = 18 9. If p + q = 5 and pq = 6, find p3 + q3 (p + q)3 = p3 + 3pq (p + q) + q 3 53 = p3 + 3.6 (5) + q 3 125 = p 3 + 90 + q 3 p3 + q3 = 125 β 90 p3 + q3 = 35 10. If a β b = 3 and ab = 10, find a 3 β b3 Given a β b = 3 and ab = 10 (a β b)3 = a3 β b3 - 3ab (a β b) 33 = a3 β b3 β 3.10 (3) 27 = a3 β b3 β 90 a3 β b3 = 27 + 90 a3 β b3 = 117 11. If a2 + π ππ = 20 and a 3 + a3 + 1 a3 π ππ = 30, find a + 1 1 a a2 = (a + ) (a2 + 1 30 = (a + ) = (20 β 1) a 1 30 = (a + ) x 19 a 30 19 = a+ 1 a 1 ππ a ππ a+ = 1 -ax ) a π π EXERCISE 3.1.4 1. Expand the following: (i) (a + b + 2c)2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a = a, b = b and c = 2c (a + b + 2c)2 = a2 + b2 + (2c)2 + 2ab + 2b (2c) + 2 (2c)a = a2 + b2 + 4c 2 + 2ab + 4bc + 4ca (ii) (x + y + 3z)2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a = x, b = y and c = 3z (x + y + 3z)2 = x2 + y2 + (3z)2 + 2.x.y + 2y(3z) + 2.(3z)x = x2 + y2 + 9z2 + 2xy + 6yz + 6zx (iii) (p + q - 2r)2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a = p, b = q and c = -2r (p + q - 2r)2 = p2 + q2 + (-2r)2 + 2.p.q + 2q(-2r) +2(-2r)p = p2 + q2 + 4r2 + 2pq β 4pr β 4pr (iv) a b 2 2 ( + c + )2 2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a b 2 2 a= ,b= a b 2 2 ( + c and c = 2 c a b c a b 2 2 2 2 2 2 + )2 =( )2 + ( )2 + ( )2 + 2 ( ) ( c a 2 2 2( )( = (v) a2 4 + b2 4 + c2 4 b c 2 2 )+2( )( )+ ) + ab 2 + bc 2 + ca 2 (x2 + y2 + z)2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a = x2, b = y2 and c = z (x2 + y2 + z)2 = (x2)2 + (y2)2 + (z)2 + 2x2y2 + 2y2z +2zx2 = x4 + y4 + z2 + 2x2y2 + 2y2z +2zx2 (vi) (m β 3 - 1 m )2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a = m, b = -3 and c = (m β 3 - 1 m )2 = m2 + (-3)2 + ( = m2 + 9 + 1 m2 1 m 1 m )2 + 2.m(-3) + 2(-3)(- - 6m + 6 m -2 1 m ) + 2 (- 1 m )m = m2 + 1 m2 + 6 m β 6m + 7 (vii) (-a + b β c)2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a = βa c = βc b=b (-a + b β c)2 = (βa)2 + b2 + (βc)2 + 2(βa)b + 2b(βc) +2(βc)a = a2 + b2 + c 2 β 2ab β 2bc +2ca (viii) (x + 5 + 1 2x )2 Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get a=x (x + 5 + 1 2x b=5 c= 1 2x 1 1 2x 2x )2 = x2 + 52 + ( )2 + 2.x.5 + 2.5. = x2 + 25 + = x2 + 1 4x 2 1 4x 2 5 +10x + + 1 x 5 +10x + + 26 x 1 + 2( )x 2x 2. Simplify the following: (i) (a β b + c)2 β (a β b β c)2 Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get (a β b + c)2 β (a β b β c)2 = [ a2 + (-b)2 + c 2 + 2a(-b) + 2(-b)c + 2ca ] β [a + (-b)2 + (βc)2 + 2a(βb) + 2(βb)(βc) + 2(βc)a] = a2 + b2 + c 2 β 2ab β 2bc +2ca β [a2 + b2 + c2 β 2ab + 2bc β2ca] = a2 + b2 + c 2 β 2ab β 2bc +2ca β a2 β b2 β c2 + 2ab β 2bc +2ca = 4ac β 4bc = 4c (a β b) (ii) (3x + 4y + 5)2 β (x + 5y β 4)2 Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get (3x + 4y + 5)2 β (x + 5y β 4)2 = [(3x)2 + (4y)2 + 52 + 2.3x.4y + 2.4y.5 + 2.5(3x)] β [x2 + (5y)2 + (-4)2 + 2. X.5y + 2.5y (-4) + 2 (-4). x] = 9x2 + 16y2 + 25 + 24xy + 40y + 30x β [x2 + 25y2 + 16 + 10xy β 40y-8x = 9x2 + 16y2 + 25 + 24xy + 40y + 30x - x2 - 25y2 - 16 - 10xy + 40y + 8x = 8x2 β 9y2 β 14xy + 80y + 38x + 9 (iii) (2m β n - 3p)2 + 4mn - 6np + 12pm Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get (2m β n - 3p)2 + 4mn - 6np + 12pm = (2m)2 + (βn)2 + (-3p)2 + 2.2m(βn) + 2(n)(β3p) + 2 (-3p)(2m) + 4mn β 6np + 12pm = 4m2 + n2 + 9p2 β 4mn β 6np β 12pm + 4mn β 6np + 12pm = 4m2 + n2 + 9p2 (iv) (x + 2y + 3z + r)2 + (x + 2y + 3z β r)2 Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get a = (x + 2y) b = 3z c=r (x + 2y + 3z + r)2 + (x + 2y + 3z β r)2 = (x + 2y)2 + (3z)2 + r2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y)2 + (3z)2 β (r)2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y) = 2(x + 2y)2 + 9z2 + r2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 + 6 (x + 2y) z - 6zr β 2r (x + 2y) = 2(x2 + 2.x.2y +4y2) + 18z2 + 2r2 + 12 (x + 2y)z = 2x2 + 8xy + 8y2 + 18z2 + 2r2 +12xz + 24 yz = 2x2 + 8y2 + 18z2 + 2r2 + 8xy +12xy + 24 yz 3. If a + b + c = 12 and a 2 + b2 + c 2 = 50, find ab + bc + ca. Given a + b + c = 12 squaring both sides (a + b + c)2 = 122 a2 + b2 + c 2 + 2ab + 2bc + 2ca = 144 Given a2 + b2 + c 2 = 50 50 + 2ab + 2bc + 2ca = 144 2(ab + bc + ca) = 144 β 50 2(ab + bc + ca) = 94 ab + bc + ca = 94 2 ab + bc + ca = 47 4. If a2 + b2 + c 2 = 35 and ab + bc + ca = 23, find all possible values of a + b +c. Given a2 + b2 + c 2 = 35 and ab + bc + ca = 23, (a + b +c)2 = a2 + b2 + c 2 + 2ab + 2bc + 2ca = a2 + b2 + c2 + 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 β 24yz β 16zx as the square of a trinomial Using and comparing the coefficient of (a +b+c)2 = a2 + b2 + c 2 + 2ab + 2bc +2ca we get 4x + 9y + 16z + 12xy β 24yz β 16zx = (2x)2 + (3y)2 + (β4z2) + 2.2x.3y.(β4z) + 2.(β4z) + 2.(β4z).2x = (2x + 3y β 4z)2 6. If x and y are real numbers and satisfy the equation (2x + 3y β 4z)2 + (5x β y - 4)2 = 0 find x, y. [Hint: If a, b are real numbers such that a2 + b2 = 0, then a = b = 0.] Given if a2 + b2 = 0, then a = b = 0 (2x + 3y β 4z)2 = 0 and (5x β y - 4)2 = 0 2x + 3y = 5 x 1 5x β y = 4 x 3 2x + 3y = 5 2x + 3y = 5 21 + 3y = 5 15x β 3y = 12 2 + 3y = 5 17x 3y = 5 β 2 = 17 x= 17 3y = 3 17 3 x=1 y= =1 3 x=1 y=1 EXERCISE 3.1.5 I. If a + b + c = 0, prove the following: (i) (b + c) (b β 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 β c) + a (a + 2b) = (-a) (b β c) + a (a + b + b) = -ab + ac + a (-c + b) = -ab + ab - ac + ac = 0 = R.H.S (ii) a (a2 β bc) + b (b2 β c) + c (c2 β ab) = 0 L.H.S = a (a2 β bc) + b (b 2 β c) + c (c2 β ab) = a3 β abc + b 3 - abc + c 3 β abc = a3 + b3 + c 3 β 3abc We know that if a + b + c = 0 then a3 + b3 + c 3 = 3abc Hence we have = 3abc β 3abc = 0 = R.H.S (iii) a (b2 + c 2) + b (c 2 + a2) + c (a 2 + b2) = β3abc L.H.S = a (b 2 + c 2) + b (c 2 + a2) + c (a2 + b2) = ab2 + ac 2+ bc 2 + ba2 + ca2 +cb 2 = ab2 + ba2 + b2c + bc 2 + ac 2 + a2c = ab (a + b) + bc (b + c) + ac (a + c) = ab (βc) + bc (βa) + ac (βb) = βabc β abc β abc = β3abc = R.H.S [a + b + c =0, a + b = -c, b + c = -a, c + a = -b] (iv) (ab + bc + ca)2 = a2b2 + b2c2 + c 2a2 L.H.S = (ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 +2ab.bc + 2bc. ca + 2ca.ab = a2b2 + b2c2 + c2a2 + 2ab 2c + 2bc 2a + 2ca2b = a2b2 + b2c2 + c2a2 + 2abc + (0) = a2b2 + b2c2 + c 2a2 = R.H.S (v) a2 β bc = b2 β ca = c 2 β ab = - (ab + bc + ca) a. a β bc b2 β ca c2 β ab a (βb β c) β bc b. b β ca c. c β ab b(βc β a) β ca c(βa β b) β ab βab β ac β ba βbc β ab β ca βac β bc β ab β (ab + bc + ca) β (ab + bc + ca) β (ab + bc + ca) β¦β¦β¦ (1) β¦β¦. (2) From equation (1) (2) and (3) a2 β bc = b 2 β ca = c 2 β ab = - (ab + bc + ca) (vi) 2a2 + bc = (a β b) (a βc) L.H.S = 2a2 + bc = a2 + a2 + bc = a2 + a x a + bc = a2 + a (- b β c) + bc = a2 β ab β ac + bc = a (a β b) β c (a - b) = (a β b) (a βc) = R.H.S β¦β¦β¦(3) (vii) (a + b) (a β b) + ca - cb = 0 We have a+b+c=0 a+b=c L.H.S = (a + b) (a β b) + ac β cb = βc (a β b) + ac β cd = β ca + bc + ac β cb = 0 = R.H.S (viii) a2 + b2 + c 2 = -2(ab + bc + ca) We have a+b+c=0 Squaring we get (a + b + c)2 = 0 a2 + b2 + c 2 + 2ab + 2bc +2ca = 0 a2 + b2 + c 2 = β 2ab β 2bc β 2ca a2 + b2 + c 2 = β 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) ππ ππ + ππ ππ ππ + ππ L.H.S = = = =3 a2 + bc b2 ca + c2 ab a 2 .a+b 2 .b +c 2 .c abc a 3 +b 3 +c 3 abc β¦β¦β¦. (1) We have a + b + c = 0, a + b = βc Cubing we get (a + b)3 = (βc)3 a3 + b3 + 3ab + (a + b) = βc3 a3 + b3 β 3ab = βc 3 a3 + b3 + c 3 = 3abc Substituting (2) in (1) L.H.S = 3abc abc =3 β¦β¦β¦. (2) (ii) ( π+π π + π+π π + π+π π )( π π+π + π + π+π π ) π+π Whenever b + c β 0, c + a β 0, a + b β 0 We have a + b + c =0 a + b = βc b + c = βa c + a = βb L.H.S = ( =( a +b c βc + c b+c + a βa a + c+a + b βa b )( )( b βb b c+a + + c βc c a +b + a βa + a b +c ) ) = (-1-1-1) (-1-1-1) = (-3) (-3) = 9 = R.H.S (iii) ππ πππ +ππ + ππ πππ ππ + ππ ππ π ππ = 1, provided the denominators do not become 0. L.H.S = = = a2 2a 2 +bc + b2 2b 2 ca a2 aβb (aβc) a2 a βb (aβc) + - + c2 2c 2 ab b2 b βc (b βa) b2 a βb (b βc) + + c2 cβa (cβb) c2 aβc (bβc) = = = a 2 bβc β b 2 aβc + c 2 aβb aβb (bβc)(aβc) a 2 bβ a 2 c β b 2 a + b 2 c + c 2 aβ c 2 b ab β b 2 β ac +bc (a βc) a 2 bβ a 2 c β b 2 a + b 2 c + c 2 aβ c 2 b a 2 b β ab 2 β a 2 c + abc β abc + b 2 c + ac 2 = 1 = R.H.S 3. If a + b + c = 0, prove that b2 β 4ac is a square. We have a + b + c = 0 b = - (a + c) Squaring on both sides b2 = [- (a + c)]2 = (a + c)2 b2 = a2 + c 2 + 2ac Subtracting 4ac on both sides b2 β 4ac = a2 + c2 + 2ac β 4ac = a2 - 2ac + c 2 b2 β 4ac = (a - c)2 We find that b 2 β 4ac is the square of (a β c) 4. If a, b, c are real numbers such that a + b + c = 2s, prove the following: (i) s (s β a) + s (s - b) + s (s β c) = s 2 L.H.S. = s (s β a) + s (s - b) + s (s β c) = s 2 β as + s 2 β bs + s 2 β cs = 3s 2 β as β bs β cs = 3s 2 β s (a + b + c) = 3s 2 β s (2s) (a + b + c = 2s) = 3s 2 β 2s 2 = s 2 = R.H.S. (ii) s 2 (s β a)2 + s (s - b)2 + s (s β c)2 = a2 + b2 + c 2 L.H.S. = s 2 (s β a)2 + s (s - b)2 + s (s β c)2 = s 2 + s 2 + a2 β 2sa + s 2 + b2 + s 2 + c 2 β 2as β 2bs β 2cs = 4s 2 + a2 + b2 + c 2 β 2s β 2bs β 2cs = 4s 2 + a2 + b2 + c 2 β 2s (a + b + c) = 4s 2 + a2 + b2 + c 2 β 2as (2s) = 4s 2 + a2 + b2 + c 2 β 4s 2 = a2 + b2 + c 2 = R.H.S. (a + b + c = 2s) (iii) (s β a) (s β b) + (s β b) (s β c) + (s β c) (s β a) + s 2 = ab + bc + ca L.H.S. = (s β a) (s β b) + (s β b) (s β c) + (s β c) (s β a) + s 2 = s 2 β as β bs + ab + s 2 β bs β cs + bc + s 2 β cs β as + ac + s 2 = 4s 2 β 2 as β 2bs β 2cs + ab + bc +ca = 4s 2 β 2s (a + b + c) + ab + bc +ca = 4s 2 β 2s (2s) + ab + bc +ca = 4s 2 β 4s 2 + ab + bc +ca = ab + bc +ca = R.H.S. (iv) a2 β b2 β c2 + 2bc = 4 (s β b) ( s β c) L.H.S = a2 β b2 β c2 + 2bc = a2 β (b2 + c2 β 2bc) = a2 β (b β c)2 = (a + b β c) [a β (b β c)] = (a + b β c) (a + b β c) = (2s β c β c) (2s β b β b) = (2s β 2c) (2s β 2b) = 2(s β c) (2) (s β b) = 4 (s β c) (s β b) = R.H.S. (a + b + c = 2s) 5. If a, b, c are real numbers, a + b + c =2s and s β a β 0, s β b β 0, s β c β 0, Prove that L.H.S = = = = = = = = = π (π¬βπ) a (sβa ) + + π (π¬βπ) b (s βb) + + π (π¬βπ) c (sβc) +2= πππ π¬βπ π¬βπ (π¬βπ) +2 a s βb sβc + b sβa s βc + c s βa sβb + 2(s βa)(s βb)(sβc) sβa s βb (Sβc) [a s 2 βbs βcs +bc +b s 2 βas βcs +ac +c s 2 βas βbs +ab +2 s 2 βas βbs +ab s βc ] s βa s βb (Sβc) as 2 βabs βacs +abc +b s 2 βabc βbcs +abc +cs 2 βacs βbcs +abc +2(s 3 βas 2 βb s 2 +abs βcs 2 +acs +bcs βabc ] s βa s βb (S βc ) [s 2 a+b+c β 2abc β2acs β 2bcs β 3abc + 2s 3 β2 as 2 β 2bs 2 + 2abs β2cs 2 + 2acs + 2bcs β2abc ] s βa s βb (S βc) [s 2 2s + abc +2s 3 β2s 2 a+b+c ] (s βa) sβb (Sβc) [2s 3 + abc +2s 3 β2s 2 2s ] (s βa) s βb (S βc) 4s 3 + abc β4s 3 (sβa ) s βb (S βc) abc (sβa ) s βb (S βc) = R.H.S. 6. If a + b + c = 0, prove that a 2 β bc = b2 β ca = c 2 β ab = (ππ + ππ +π π ) π We have a + b + c = 0 Squaring will get (a + b + c)2 = 0 a2 + b2 + c 2 + 2ab + 2bc + 2ca = 0 a2 + b2 + c 2 + 2b (a + c) + 2ca = 0 a2 + b2 + c 2 + 2b (-b) + 2ca = 0 a2 + b2 + c 2 = 2b2 β 2ca (a + c = -b) a2 + b2 + c 2 = 2 (b2 β ca) a 2 + b 2 +c 2 2 IIIly = b2 β ca β¦β¦β¦ (1) (a + b + c)2 = 0 a2 + b2 + c 2 + 2ab + 2bc + 2ca = 0 a2 + b2 + c 2 + 2ab + 2c (b + a) = 0 a2 + b2 + c 2 + 2ab + 2c (-c) = 0 (a + c = -c) a2 + b2 + c 2 = 2 (c 2 β ab) a 2 + b 2 +c 2 2 = (c 2 β ab) β¦β¦β¦ (2) Also a2 + b2 + c 2 + 2ab + 2bc + 2ca = 0 a2 + b2 + c 2 + 2a (b + c) + 2bc = 0 a2 + b2 + c 2 + 2a (-a) + 2bc = 0 a2 + b2 + c 2 = 2 (a2 β bc) β¦β¦..(3) From (1), (2) and (3) we get, 2 2 2 a β bc = b β ca = c β ab = (a 2 + b 2 +c 2 ) 2 7. If 2(a 2 + b2) = (a + b)2, prove that a = b. 2a2 + 2b 2 = a2 + b2 + 2ab 2a2 + 2b 2 β a2 β b2 β 2ab = 0 a2 + b2 β 2ab = 0 (a β b)2 = 0 a β b =0 a =b 8. If x2 β 3x + 1 = 0, prove that x2 + π π±π = 7. Given x2 β 3x + 1 = 0 x2 + 1 = 3x 1 x + = 3 (dividing both sides by x) x Squaring both sides we get 1 1 x x2 (x + )2 = 32 = x2 + x2 + 1 x2 x2 + 1 x2 1 + 2x. = 9 x + 2= 9 =9β2=7 ADDITIONAL PROBLEMS ON βMULTIPLICATION OF POLYNOMIALSβ I. Find the following products: (i) (2a + 3b) (4a 2 β 6ab + 9b2) = 8a3 + 12a2b β 12ab2 β 18ab2 + 18ab 2 + 27b 3 = 8a3 + 27b3 (ii) (3x + 4y) (9x2 β 12xy +16y2) = 27x3 + 36x2y - 36x2y - 48xy2 + 48xy2 + 64y3 = 27x3 + 64y3 (iii) (5x β 2y) (25x2 + 10xy + 4y2) = 125x3 β 50x2y + 50x2y β 20xy2 + 20xy2 β 8y3 = 125x3 β 8y3 (iv) (a3 β 2) (a 3 + 2a3 + 4) = a9 β 2a6 + 2a6 β 4a3 + 4a3 β 8 = a9 β 8 2. By which factor should the following get multiplied to be in the form a3 + b3. (i) (2x + 1) We have a3 + b3 = (a + b) (a2 β ab +b2) a = 2x b=1 (2x)3 + 1 = (2x + 1) [(2x)2 β 2x.1 + 12] 8x + 1 = (2x + 1) (4x2 β 2x + 1) We have to multiply (2x + 1) with (4x2 β 2x + 1) (ii) 4x2 β 6x + 9 We have a3 + b3 = (a + b) (a2 β ab +b2) a = 2x b=3 (2x)3 + 33 = (2x + 3) (4x2 β 6x + 9) (2x + 3) should be multiplied (iii) 9a2 β 15a +25 We have a3 + b3 = (a + b) (a2 β ab +b2) a = 3a b=5 (3a)3 + 5 = (3a + 5) (9a β 15a + 25) Hence (3a + 5) should be multiplied. (iv) 4a + 3 We have a3 + b3 = (a + b) (a2 β ab +b2) a = 4a b=3 (4a)3 + 3 = (4a + 3) [(4a)2 β 4a.3 + 32) = (4a + 3) (16a2 β 12a + 9) Hence (16a2 β 12a + 9) should be multiplied. 3. By which factor should the following get multiplied to be in the form a3 - b3. (i) (5a β 3) We have a3 - b3 = (a - b) (a2 + ab +b 2) a = 5a b=3 (5a)3 β 33 = (5a β 3 ) [(5a2) + (5a) 3 + 32] 125a3 β 27 = (5a β 3) (25a2 + 15a + 9) (25a2 + 15a + 9) should be multiplied. (ii) `16a2 + 20a + 25 We have a3 - b3 = (a - b) (a2 + ab +b 2) a = 4a b=5 (4a)3 β 53 = (4a β 5 ) (4a2 + 20a + 25) (4a β 5) should be multiplied. (iii) 3x β 2y We have a3 - b3 = (a - b) (a2 + ab +b 2) a = 3x b = 2y (3x)3 β (2y)3 = (3x β 2y) [(3x)2 + 3x.2y + (2y)2] 27x3 β 8y3 = (3x β 2y) (9x2 + 6xy + 4y2) (9x2 + 6xy + 4y2) should be multiplied. (iv) 16x2 β 20x + 25 We have a3 - b3 = (a - b) (a2 + ab +b 2) a = 4x b=5 (4x)3 β 53 = (4x β 5 ) (4x2 + 20x + 25) (4x β 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 = a2 + 2ab + b 2 We get a = 100 b=3 (100 + 3)2 = 1002 + 2.100.3 +32 = 10000 + 600 + 9 = 10609 (ii) (107)2 We write 107 = 100 + 7 Using (a + b)2 = a2 + 2ab + b 2 We get a = 100 b=7 (100 + 7)2 = 1002 + 2.100.7 +72 = 10000 + 1400 + 49 = 11449 (iii) π ( 50 )2 π We write 50 1 2 = 50.5 = 50. +0.5 Using (a + b)2 = a2 + 2ab + b 2 We get a = 50 b = 0.5 (50 + 0.5)2 = 502 + 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 = a2 + 2ab + b 2 We get a = 1000 b = -2 (1000 - 2)2 = 10002 - 2.1000.2 +22 = 1000000 β 4000 + 4 = 996004 (v) 107 x 93 We write 107 = 100 + 7 and 93 = 100 -7 Using (a + b) (a β b) = a2 + b2 We get a = 100 b=7 (100 + 7) (100 β 7) = 1002 β 72 = 10000 β 49 = 9951 (vi) 1008 x 992 We write 1008 = 1008 + 8 and 992 = 1000 - 8 Using (a + b) (a β b) = a2 + b2 We get a = 1000 (1000 + 8) (1000 β 8) = 10002 β 82 = 100000 β 64 = 99936 b=8 (vii) (101)3 We write 101 = 100 + 1 Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 We get a = 100 b=1 (100 + 1)3 = 1003 + 3(100)2.1 + 3(100).12 + 13 = 1000000 + 30000 + 300 +1 = 10303001 (viii) (1002)3 We write 1002 = 1000 + 2 Using (a + b)3 = a3 + 3a2b + 3ab 2 + b3 We get a = 1000 b=2 (1000 + 2)3 = 10003 + 3(1000)2.2 + 3(1000).22 + 23 = 1000000000 + 6000000 + 12000 + 8 = 1006012008 5. If x + y = 7 and xy = 12, find x2 + y2 and x2 + y2. Given x + y = 7 and xy = 12 We take x+y=7 Squaring both sides (x + y)2 = 72 x2 + y2 + 2xy = 49 x2 + y2 + 2.12 = 49 x2 + y2 = 49 β 24 x2 + y2 = 25 IIIly x + y = 7 Cubing both sides (x + y)3 = 73 x3 + y3 + 3xy(x + y) = 343 x3 + y3 + (3 x 12 x 7) = 343 x3 + y3 = 343 β 252 x3 + y3 = 91 6. π π π± π±π If x + = 3, find x2 + and x4 + π π±π 1 x+ =3 x Squaring both sides (x + x2 + x2 + x2 + 1 2 ) = x 1 x2 1 x2 1 x2 32 1 + 2.x. = 9 x = 9β 2 =7 Squaring again (x2 + x4 + x4 + x4 + 1 x2 1 x4 1 x4 1 x4 )2 = 72 + 2.x2. 1 x2 = 49 = 49 β 2 = 74 7. If a β b = 2 and ab = 15, find a 3 β b3. aβb =2 Cubing both sides (a β b)3 = 23 a3 β b3 β 3ab (a β b) = 8 a3 β b3 β 3 x 15(2) = 8 a3 β b3 = 8 + 90 a3 β b3 = 98 8. If a + b + c = 2s then prove that (s β a)3 + (s β b)3 + 3c (s β a) (s β b) = c 3 Let s β a = x and s β b = y Now L.H.S. will be x+y x3 + y3 + 3cxy s β a + s βb x3 + y3 + 3(x + y) xy (x + y = c) (x + y)3 c3 2s β a - b a+b+cβaβb c 9. If a + b + c = 2s, prove that 16s (s β a) (s β b) (s β c) = 2a 2b2 + 2b2c2 + 2c 2a2 β a2 β a4 β b4 β c4 Given a + b + c = 2s L.H.S. = 16s (s β a) (s β b) (s β c) = 2s 2(s β a) 2(s β b) 2(s β c) = 2s (2s β 2a) (2s β 2b) (2s β 2c) = (a + b + c) (a + b + c β 2a) (a + b + c β 2b) (a + b + c β 2c) = (a + b + c) (b + c β a) (a + c β b) (a + b β c) = (ab + b2 + bc + ac + bc + c 2 β a2 β ab β ac) = (a2 + ac β ab β ab + bc β b2 β ac β c2 + bc) = (b2 + c2 β a2 + bc ) (a2 β b2 β c + 2bc) = a2b2 + a2c2 β a4 + 2a2 bc β b4 β b2c 2 + a2b2 β 2b3c β c 2b2 β c4 + a2c2 - 2bc 3 + 2b3c + 2bc 3 β 2a3bc + 4 b2c2 = 2a2b2 + 2 b2c2 + 2a2c2 β a4 β b4 β c4 = R.H.S. 10. If a2 + b2 = c 2, prove that (a + b + c) (b + c β a) (c + a β b) (a + b β c) = 4a2b2 L.H.S = (a + b + c) (b + c β a) (c + a β b) (a + b β c) Re arranging (a + b + c) (a + b β c) (c + b β a) c β (b β a) (a + b)2 β c2 c2 β (b β a2) a2 + b2 + 2ab β c 2 c2 β (a2 + b2 β 2ab) (a2 + b2) + (2ab β c2) c2 β (a2 + b2) + 2ab (c 2+ 2ab β c 2) c2 β c2 + 2ab (2ab) (2ab) 4a2b2 11. If x + y = a and xy = b, prove that (1 + x2) (1 + y2) = a2 + (1 + b2). ` L.H.S =(1 + x2) (1 + y2) = 1 + x2 + y2 + x2y2 Add and subs tract 2xy = 1 + x2 + y2 + 2xy β 2xy + x2y2 = 1 + (x + y)2 β 2xy + (xy)2 = 1 + a2 β 2b + b 2 [x + y = a, xy = b] = a2 + (1 β 2b + b 2) = a2 + (1 β b)2 = R.H.S [(a - b)2 = a2 β 2ab + b 2] π π π± π±π 12.If x β = 4, show that x3 + 6x2 + - π π±π = 184. 1 Given x β = 4 x Squaring both sides 1 (x β )2 = 42 x x2 + x2 + x2 + 1 x2 1 x2 1 x2 1 - 2x. = 16 x = 16 + 2 = 18 β¦β¦β¦..(1) 1 Again take x β = 4 x Cubing on both sides 1 (x β )3 = 43 x x3 β x3 β x3 β x3 β 1 1 x3 1 x3 1 x3 1 1 x x β 3x. (x β ) = 64 x3 β 3(4) = 64 = 64 + 12 β¦β¦β¦β¦..(2) = 76 Now consider L.H.S x3 + 6x2 + = x3 β 1 x3 6 x2 - 1 x3 + 6 (x2 + = 76 + 6 x 18 = 76 + 108 = 184 R.H.S 1 x2 ) 13. If xy(x + y) = 1, prove that π π± π π²π β x3 β y3 = 3 Given xy(x + y) = 1 By dividing the equation by xy x+y= 1 xy Cubing the both sides 1 (x + y)3 = ( )3 xy x3 + y3 + 3xy(x + y) = x3 + y3 + 3(1) = 1 x 3 y3 1 x 3 y3 β x3 + y3 = 3 1 x 3 y3 [xy (x + y) = 1] 14.Suppose a, b, are the sides of a triangle such that 2s = a + b + c. Prove that L.H.S = = = = = = = = = ππ β ππ + πππ β π π π = ππ + πππ + π π β π π¬βπ (π¬βπ) π¬( π¬βπ) a 2 β b 2 + 2bc β c 2 b 2 + 2bc + c 2 β a 2 a 2 β ( b 2 β 2bc + c 2 ) (b 2 + 2bc + c 2 ) β a 2 a 2 β (b β c)2 ( b + c 2 ) β a2 a + bβc [aβ b β c ] [ b + c + a] [ b + c β a] [a2 β b2 = (a + b) (a β b)] a + b β c (a β b+c) a + b + c (b +c βa) a + b + c β c βc (a + b + c β b βb ) a + b + c (a + b + c β a βa ) 2s β 2c (2s β 2b) 2s (2s β 2a) 4 s β c (s β b ) 4s (s β a) s β b (s β c) s (s β a ) = R.H.S 15.Suppose a, b, are the sides of a triangle such that 2s = a + b + c. Prove that `L.H.S. = = = = = π + π¬ βπ π π¬ βπ + π π¬ βπ π π¬ βπ + + π π¬β π π π¬ βπ β β π π¬ = πππ π¬ π¬ β π π¬ β π (π¬ β π) π π¬ s s βb s βc + s s βa s βc + s s βa s βb β [ s βa sβb s βc ] s s βa s βb (S βc) s s 2 βbs βcs +bc +s s 2 βas βcs +ac +s s 2 βas βbs +ab s s βa s βb (S βc) s 3 β bs 2 βcs 2 + bcs + s 3 β a s 2 β cs 2 + acs + s 3 β cs 2 β bs 2 + abs s s βa s βb (Sβc) 3s 3 β2as 2 β2b s 2 β2bc + abs + bcs + acs β s 3 + a s 2 + bs 2 β abs + cs 2 β acs β bcs +abc s sβa s βb (Sβc) = [2s 3 β as 2 β bs 2 β cs 2 + abc ] = [2s 3 β s 2 a + b + c + abc ] = [2s 3 β s 2 2s + abc ] = = 2 s 3 β 2s 3 + abc s (s βa) s βb (S βc) abc s (s βa) s βb (S βc) = RHS 1 s(sβa ) s βb (S βc) 1 s (s βa) s βb (S βc) 1 s (sβa) s βb (Sβc) (a + b + c = 2s) EXTRA QUESTIONS I. Expand using appropriate identity: (i) (3a + 5b)2 Using (a + b)2 = a2 + 2ab + b 2 we get a = 3a b = 5b (3a + 5b)2 = (3a)2 + 2.3a.5b + (5b)2 = 9a2 + 30ab + 25b2 (ii) ( π π π±+ π π π²)2 Using (a + b)2 = a2 + 2ab + b 2 we get a= ( 1 2 1 2 2 x x+ b= y 3 2 3 = (iii) 1 x 2y 2 2 3 y)2 = ( x)2 + 2. . x2 4 2 4y 2 3 3 + xy + 2y + ( )2 3 (2a + 3b + 4c)2 Using (a + b + c)2 = a2 + b2 + 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 = 4a2 + 9b2 + 16c 2 + 12ab + 24bc + 16ca (4a β 3b)2 (iv) Using (a β b)2 = a2 β 2ab + b2 we get a = 4a b = 3b (4a β 3b)2 = (4a)2 β 2.4a.3b + (3b)2 = 16a2 β 24ab + 9b2 (3a β 2b + 5c)2 (v) Using (a + b + c)2 = a2 + b2 + c 2 + 2ab + 2bc + 2ca we get a = 3a b = -2b c = 5c (3a β 2b + 5c)2 = (3a)2 + (β2b)2 + (5c)2 + 2(3a) (β2b) + 2(β2b) (5c) + 2(5c) (3a) = 9a2 + 4b2 + 25c 2 β 12ab β 20bc + 30c II. If (x β π ππ± π (i) x2 + (i) (x β 1 2x ) = 3, find the valve of (ii) x4 + ππ± π π πππ± π )=3 Squaring on both sides (x β x2 + x2 + 1 2x 1 4x )2 = 32 β 2.x. 2 1 4x 2 1 2x =9 = 9 + 1 = 10 (iii) x3 β π ππ± π (ii) x2 + 1 = 10 4x 2 Squaring on both sides (x2 + x4 + x4 + x4 + (iii) (x β 1 4x 2 1 16 x 4 1 1 16 x 4 2x + 2.x2 x 1 = 100 4x 2 1 16 x 4 1 )2 = 102 + = 100 2 = 100 - 1 2 = 99 1 2 )=3 Cubing on both sides (x β 1 2x (x3 β 1 8x 3 = x3 β x3 β )3 = 33 - 3.x x 1 8x 3 1 8x 3 1 2x = (x β 1 2x ) = 27 3 = (3) = 27 2 9 54 + 9 2 2 = 27 + = = 63 2 π If a2 + 3. (i) a + (i) = 14, find the values of ππ π (ii) a β π a2 + 1 a2 π (iii) a2 β π = 14 Additing 2 on both sides (a2 + 1 a2 + 2) = 14 + 2 1 (a2 + 2 )2 = 16 a 1 a + = 12 = ±4 a (ii) a2 + 1 a2 = 14 Subtracting 2 on both sides 1 (a2 + a2 β 2) = 14 β 2 1 (a β )2 = 12 a 1 (a β ) = 12 = ±2 3 a (iii) a2 β 1 a2 1 1 a a = (a β ) (a β ) = (±4) (±2 3) = (±8 3) π ππ 4. If (3a + 4b) = 16 and ab = 4 find the value of 9a 2 + 16b2 We have (3a + 4b) = 16 Squaring on both sides (3a + 4b)2 = 162 9a2 + 2.3a.4b + 16b 2 =256 9a2 + 16b 2 = 256 β 24ab = 256 β 24 x 4 = 256 -96 = 160 5. If a2 β 4a β 1 = 0 2a β 0, find the values of π (ii) a β (i) (a + ) π We have a2 β 4a β 1 = 0 a2 β 4a = 1 Adding 4 on both sides a2 β 4a + 4 = 1 + 4 (a + 2)2 = 5 a+2= 5 a= 5β2 1 a = 1 5β2 π π (i) 1 a+ = 5β2+ a = = = = (ii) 5β2 ( 5β2)2 + 1 5β2 5 + 4 β 2.2 5 + 1 5β2 10 β 4 5 5β2 1 a = = = = 5β2 ( 5β2)2 + 1 aβ = 5β2= 1 1 5β2 ( 5β2)2 β 1 5β2 5+4β2 5β 2 β 1 5β2 9 β 4 5β1 5β2 9β4 5 5β2 4(2 β 5 ) β(2 β 5 ) =β4 6. If a + 2b + 3c = 0, show that a 3 + 8b3 +27c 3 β 18abc We have a + 2b + 3c = 0 a + 2b = β3c Cubing on both sides (a + 2b)3 = (β3c)3 a3 + 8b3 + 3a (2b) (a+2b) = -27c 3 a3 + 8b3 + 6ab (-3c) = -27c 3 a3 + 8b3 + 27c 3 = 18abc π π π π 7. If = prove that (a + b + c) (a β b + c) = a 2 + b2 + c 2 a b b c [Hint: Let = = k. then b = ck, a = bk, b = ck and a = ck2] L.H.S = (a + b + c) (a β b + c) = (ck2 + ck + c) (ck2 - ck + c) = c (k2 + k + 1) c (k2 - k + 1) = c 2 (k2 + 1 + k) (k2 +1 - k) = c 2 [(k2 β 1)2 β k] = c 2 [k4 + 2k2 β 1 β k2] = c 2 [k4 + k2 + 1] β¦β¦β¦. (1) R.H.S = a2 + b2 + c 2 = (ck2)2 + (ck)2 + c 2 = c 2k4 + c 2k2 + c2 = c 2 (k4 + k2 + 1) β¦β¦β¦.. (2) From (1) and (2) we get LHS = RHS 8. Evaluate (i) (968)2 β (32)2 (i) (ii) (98.7)2 β (1.3)2 (968)2 β (32)2 Using (a2 β b2) = (a + b) (a β b) we get a = 968 and b = 32 (968)2 β (32)2 = (968 + 32) (968 β b) = 1000 x 936 = 936000 (ii) (98.7)2 β (1.3)2 Using (a2 β b2) = (a + b) (a β b) we get a = 98.7 and b = 1.3 (98.7)2 β (1.3)2 = (98.7 + 1.3) (98.7 β 1.3) = 100 x 97.4 = 9740 9. If x + y = 8 and xy = 12, find x4 + y4. Take x + y = 8 Squares we get (x + y)2 = 82 x2 + y2 + 2xy = 64 x2 + y2 + 2 x 12 = 64 x2 + y2 = 64 β 24 x2 + y2 = 40 Squaring both sides we get (x2 + y2)2 = 402 x4 + y4 + 2x2y2 = 1600 x4 - y4 + 2(xy)2 = 1600 x4 + y4 + 2(12)2 = 1600 x4 + y4 + 2 x 144 = 1600 x4 + y4 = 1600 β 288 x4 + y4 = 1312 10. If a + b = 8 and ab = 15, find a3 + b3. Take a + b = 8 Cubing both sides (a + b)3 = 83 a3 + b3 + 3ab (a + b) = 512 a3 + b3 + 3 x 15(8) = 512 a3 + b3 + 360 = 512 a3 + b3 = 512 β 360 a3 + b3 = 152 π π π± π±π 11. If x β = 3, find x3 β 1 Take x β = 3 x Cubing on both sides 1 (x β )3 = 33 x 1 1 1 x x x x3 + ( )3 + 3x + (x β ) = 27 1 x3 + x3 1 x3 + x3 1 x3 + x3 + 3(3) = 27 = 27 β 9 = 18 π π π± π±π 12. If x β = 5, find x3 β 1 Take x β = 5 x Cubing on both sides 1 (x β )3 = 53 x 1 1 1 x x x x3 β ( )3 β 3x + (x β ) = 125 x3 β x3 β x3 β x3 β 1 β 3(5) = 125 x3 1 x3 β 15 = 125 1 = 125 + 15 x3 1 = 140 x3 13. Suppose x, y, z are non-zero real numbers such that x + y + z = 0 (π± + π²)π π±π² (π² + π³) π + (π³ + π±)π + π²π³ = 3. π³π± We have x + y = -z, y + z = -x, and z +x = -y LHS = = = = = (x + y) 2 xy (βz)2 xy z2 xy + + x2 xy + (y + z) 2 (βx) 2 zy + z 3 + x3 + y 3 xyz 3 xyz xyz = 3 RHS yz + + (z + x) 2 zx (βy)2 zx y2 xy (if x + y + z = 0 then x3 + y3 + z3 = 3xyz) 14. If a + b + c = 2s prove that s(s β a) + (s β b) + (s β c) = s 2 LHS = s(s β a) + (s β b) + (s β c) s 2 β sa + s 2 β sb + s 2 - sc 3s 2 β s (a + b + c) 3s 2 β s (2s) s 2 = RHS π π π± π±π 15. If x + = 3, find x2 + = 3β5 1 x+ =3 x Squaring on both sides 1 (x + )2 = 32 x x2 + x2 + x2 + 1 x2 1 x2 1 x2 1 + 2 x =9 x = 9β 2 β¦β¦β¦..(1) =7 1 1 1 x x x Now (x2 β )2 = (x β )2 β 4x x 1 (x β )2 = 32 β 4 x 1 (x β )2 = 5 x 1 xβ = 5 x β¦β¦β¦β¦ (2) Now x2 β 12 x 1 1 x x = (x + ) (x β ) From (1) and (2) x2 + π π±π = 3 5 = R.H.S Important formulae 1. (a + b)2 = a2 + 2ab + b 2 2. (a β b)2 = a2 β 2ab + b 2 3. (a + b) (a β b) = a2 β b2 4. (x + a) (x + b) = x2 + x (a + b) + ab 5. (x + a) (x+ b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc 6. (a + b)3 = a3 + b3 + 3ab (a + b) 7. (a + b)3 = a3 β b3 β 3ab (a β b) 8. a3 + b3 = (a + b) (a2 β ab + b 2) 9. a3 β b3 = ( a β b) (a2 + ab + b 2) 10. (a + b + c)2 = a2 + b2 + 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 β 4) 90° = (2 x 6 β 4) x 90° = (12 β 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 β 4) 90° = (2 x 8 β 4) x 90° = (16 β 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 β 4) 90° = (2 x 5 β 4) x 90° = (10 β 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 β 4) 90° = (2 x 9 β 4) x 90° = (18 β 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 β 4) 90° = (2 x 10 β 4) x 90° = (20 β 4) 90 = 16 x 90 = 1440° Sum of the exterior angles = 360° 2. How many sides dose a polygon have if the sum of the interior angle is (i) 540° (ii) 900° (iv) 7 straight angles (iii) 1440° (v) 8 straight angles Solution: (i) Sum of the interior angles = (2n β 4) 90° = 540° β 2n β 4 = 540 90 =6 β 2n = 6 + 4 = 10 n= 10 2 =5 The polygon has 5 sides (ii) Sum of the interior angles = (2n β 4) 90° = 900° β 2n β 4 = 900 90 = 10 β 2n = 10 + 4 = 14 β n= 14 2 =7 The polygon has 7 sides (iii) The Sum of the interior angles = (2n β 4) 90° = 1440° β 2n β 4 = 1440 90 = 16 β 2n = 16 + 4 = 20 β n= 20 2 = 10 The polygon has 10 sides (iv) Sum of the interior angles = (2n β 4) 90° = 7 straight angles = 7 x 180 β (2n β 4) = β 2n β 4 = 7 x 180 90 7 x 180 90 = 7 x 2 = 14 β 2n = 14 + 4 = 18 β n= 18 2 =9 The polygon has 9 sides (v) Sum of the interior angles = (2n β 4) 90° = 8 right angles β (2n β 4) 90° = 8 x 90° β 2n β 4 = 8 x 90 90 =8 β 2n = 8 + 4 = 12 β n= 12 2 =6 The polygon has 6 sides 3. Find the measure of each exterior angle of a regular polygon with sides: (i) 40 (ii) 30 (iii) 20 (iv) 18 (v) 16 (vi) 2x (vii) (2a +4b) Solutions: (i) Number of sides n = 40 Measure of each exterior angle = ( (ii) 40 = 9° 360 n )°= 360 30 = 12° 360 n )°= 360 20 = 18° Number of sides n = 18 Measure of each exterior angle = ( (v) 360 Number of sides n = 20 Measure of each exterior angle = ( (iv) n )°= Number of sides n = 30 Measure of each exterior angle = ( (iii) 360 360 n )°= 360 18 = 20° Number of sides n = 16 Measure of each exterior angle = ( 360 n )°= 360 16 = 22.5° (vi) Number of sides n = 2x Measure of each exterior angle = ( 360 )°= n 360 2x =( πππ π± )° (vii) Number of sides n = 2a + 4b Measure of each exterior angle = ( 360 )°= n 360 2a + 4b =( πππ π + ππ )° 4. Find the numbers of sides of regular polygon. If each exterior angle measures. (i) 10° (ii) 20° (v) 45° (iii) 30° (vi) 60° (iv) 40° (vii) 72° (viii) 120° Solution: (i) Measure of each exterior angle = x° = 10° Number of sides = (ii) 360 x = 360 10 = 35 Measure of each exterior angle = x° = 20° Number of sides = 360 x = 360 20 = 18 (iii) Measure of each exterior angle = x° = 30° Number of sides = (iv) 360 30 = 12 360 x = 360 40 =9 Measure of each exterior angle = x° = 45° Number of sides = (vi) x = Measure of each exterior angle = x° = 40° Number of sides = (v) 360 360 x = 360 45 =8 Measure of each exterior angle = x° = 60° Number of sides = 360 x = 360 60 =6 (vii) Measure of each exterior angle = x° = 72° Number of sides = 360 x = 360 72 =5 (viii) Measure of each exterior angle = x° = 120° Number of sides = 360 x = 360 120 =3 5. Find the numbers of sides of a regular polygon it each exterior is equal to (i) Its adjacent interior angle. (ii) Twice its adjacent interior angle. (iii) Half its adjacent interior angle. (iv) One-third of its adjacent interior angle. Solution: (i) Exterior angle + its adjacent interior = 180° e + i = 180° e=i e + e = 180° 2e = 180° e= 180 2 = 90° Number of sides = (ii) 360 e e + i = 180° e = 2i β i= βe+ e 2 e 2 = 180° β 2e+e β 3e = 180 x 2 = 360° β e= 2 = 180° 360 3 = 120° = 360 90 =4 Number of sides = (iii) 360 e = 360 120 =3 e + i = 180° e= i 2 β i = 2e β e + 2e= 180° β 3e = 180° β e= 180 3 = 60° Number of sides = (iv) 360 e = 360 60 =6 e + i = 180° e= i 3 β i = 3e β e + 3e= 180° β 4e = 180° β e= 180 4 = 45° Number of sides = 360 e = 360 45 =8 6. find the number of sides in a regular polygon if each interior angle is (i) twice its adjacent exterior angle (ii) Four times the adjacent exterior angle. (iii) Eight times the adjacent exterior angle. (iv) Seventeen times the adjacent exterior angle. Solution: (i) e + i = 180° i = 2e β e + 2e = 180° β 3e = 180° βe= 180 3 = 60° Number of sides = (ii) 360 e = 360 60 =6 e + i = 180° i = 4e β e + 4e = 180° β 5e = 180° βe= 180 5 = 36° Number of sides = 360 e = 360 36 = 10 (iii) e + i = 180° i = 8e β e + 8e = 180° β 9e = 180° 180 βe= 9 = 20° Number of sides = (iv) 360 e = 360 20 = 18 Seventeen times the adjacent exterior angle e + i = 180° i = 17e β e + 17e = 180° β 18e = 180° βe= 180 18 = 10° Number of sides = 360 e = 360 10 = 36 7. The angles of a convex polygon are in the ratio 2 : 3 : 5 : 9 :11. Find the measure of each angle. Solution: Ratio of the angle = 2 : 3 : 5 : 9 :11. Let the angle be 2x, 3x, 5x, 9x and 11x. The polygon has five angles and therefore five sides. Sum of angle = (2n - 4)90° = (2 x 5 β 4)90° = (10 β 4)90 = 6 x 90° 2x + 3x + 5x + 9x +11x = 6 x 90° 30x = 6 x 90° x= 6 x 90 30 = 6 x 3 = 18° The angles are 2x = 2 x 18 = 36° 3x = 3 x 18 = 54° 5x = 5 x 18 = 90° 9x = 9 x 18 = 162° 11x = 11 x 18 = 198° 8. Prove that the opposite sides of a regular hexagon are parallel. Solution: E D F C A B Construction: in the regular hexagon ABCDEF, join FC. In the quadrilateral ABCF, AF = BC and FAB = CBA = 120° ABCF is an isosceles trapezium FC | | AB Similarly ED | | FC β¦β¦. (1) β¦β¦. (2) From (1) and (2) AB | | ED Similarly BC | | FE and CD | | FA ADDITIONAL PROBLEMS ON POLYGONS 1. In an equiangular polygon, the measure of each exterior angle is 25% of the measure of each interior angle. How many sides it has? Solution: e + i = 180° e= ix 25 100 1 = i 4 β i = 4e β e + 4e = 180° β 5e = 180° e= 180 5 = 36° Number of sides = 360 e = 360 36 = 10 2. In a heptagon, two of the angles 130° each and the remaining angles are equal. Find the equal angles. Solution: A heptagon has 7 sides Sum of interior angles = (2n β 4)90° = (2 x 7 β 4)90 = (14 β 4) 90 = 10 x 90 = 900° Measure of two equal angles = 130° each their sum = 130 x 2 = 260 Sum of the remaining 5 equal angles = 900 β 260 = 640 Measure of each of the 5 angles = 640 5 = 128° 3. A polygon has n sides two of its angles are right angles and each of the remaining angles is 144 degrees find the value of n. Solution: Numbers of sides = n Sum of two right angles = 90° x 2 = 180° Sum of the remaining (n - 2) angles, Each equal to 144° = (n - 2)144° Sum of interior angles = (2n β 4)90° = 180 + (n β 2)144 180n - 360 = 180 + 144n β 288 180n β 144n = 180 + 360 β 288 36n = 540 β 288 = 252° n= 252 36 =7 4. Is there a polygon which has only two types of interior angles 120° and 60°? If so how many sides such a polygon has? [Hint: if n is the number of sides and if k angles are of 60° then (n β k) angles are of 120° and (n - 2) 180 = 60k + 120(n β k) Solution: 180n β 360 = 60k + 120n β 120k 180n β 120k - 60k + 120k = 360 60n + 60k = 360 n+k=6 When k = 1 n = 5 and When k = 2 n = 4 Thus the polygon can be quadrilateral with two 60°angles and two 120° Angles. or a pentagon with one 60° angle and four 120° angles. 5. In the adjacent figure, the pentagon is such that AB = BC = CD and AE = ED. Moreover ABC = BCD = AED = 130°. Find the measures of BAE and EDC B C 130° 130° A D 130° E Solution: Construction: Join AD. ABC = BCD = 130° AB = CD ABCD In an isosceles trapezium ABC + BAD = 180° 130° + BAD = 180 BAD = 180 β 130 = 50° ADC = BAD = 60 In EAD, EA = ED EAD = EDA But EAD + EDA = 180 β 130 = 50 EAD = EDA = 50 2 = 25° BAE = BAD + DAE = 50 + 25 = 75° Similarly EDC = 75° 6. In the adjacent figure you have a star ABCDEFGHIKL. All the sides of the star are equal. What is the measure of angles at the vertices A, C, E, G and k? K E F H K D L B A Solution: All s are isosceles. x+ x + A + x + x + C + β¦β¦β¦ = 180 x 5 = 900 10x + A + C + E + G + K = 900 10x + 5y = 900 5(2x + y = 180) C 7. Prove that the perpendicular drawn from the vertex of a regular pentagon to the opposite side bisects that side, Solution: D E C A F B Join DA and DB In DEA and DCB DE = DC EA = CB DEA = DCB DEA = In DCB (SAS) DAF and DBF Hypotenuse DA = Hypotenuse DB DF = DF DAF β DBF (RHS) AF = FB (CPCT) DF bisects AB 8. For what values of m and n is it possible for the external angle of a regular m-gon to be equal to the internal angle of a regular n-gon. Check your answer. π π π§ ππ¦ [Hint: (n β 2) = ] Solution: The sum of the exterior angles of a m-gon. = 2π The measure of each exterior angle = Ο 2m The sum of the interior angles of a n-gon. = (n β 2) π The measure of each exterior angle = Now, Ο 2m 2 m = = (n β 2)Ο n (n β 2)Ο n nβ2 n 2n = mn β 2m 2n = m (n β 2) m= 2n nβ2 m and n can have all positive integral values satisfying the above condition. Such that m β₯ 3 and n β₯ 3 9. In a convex polygon, sum of all the angles except one angle is 2280°. How many sides does the polygon have? What is the measure of this exceptional angle? [Hint: if x is the exceptional angle, then (n β 2) 180 = 2280 + x, and 0 < x < 180] Solution: If the polygon has n sides, the Sum of the interior angle = (n β 2) If the exceptional angle measures x° β (n β 2)180 = 2280 + x β 2280 + x is a multiple of 180 β 120 + x = 180 Or x = 180 β 120 = 60 Thus (n β 2)180 = 2280 + 60 = 2340 n -2 = 2340 180 = 13 n = 13 + 2 = 15 And the exceptional angle = 60 12 180 2280 180 480 360 120