Oasis School Mathematics-10 147 or, 3a = 51 or, a = 51 3 ∴ a = 17. Hence, 3 numbers in A.S be (17 – d), 17 and (17+ d). Now, (17 – d)² + 17² + (17 + d)² = 939. or, 289 –34d + d² + 289 + 289 + 34d + d² = 939 or, 2d²+ 867 = 939 or, 2d² = 939 – 867 or, 2d² = 72 or, d² = 36 or, d = ± 6. when d = 6, Three terms 17 –d, 17 and 17 + d = 17 – 6, 17 and 17 + 6 = 11, 17 and 23 when d = –6 Three terms 17 – d, 17 and 17 + d = 17 – (–6), 17 and 17 – 6 = 23, 17 and 11. Example 9 In an Arithmetic sequence, The sixth term is equal to the three times the fourth term and the sum of the first 3 terms is –12, find the sum of first 20 terms. Solution: Given, t6 = 3t4 and S3 = –12, S20 =? Here, Given, t6 = 3t4 a + 5d = 3(a + 3d) a + 5d =3a + 9d a – 3a = 9d – 5d –2a = 4d a = – 2d ........................ (i) Again, Sn = n 2 [2a + (n – l)d] or, S3 = 3 2 [2(–2d) +(3–1).d] or, –12 = 3 2 [–4d +2d]
148 Oasis School Mathematics-10 or, –12 = 3 2 × (–2d) or, –12 = –3d or, –12 –3 = d ∴ d = 4. From (i) a = –2d = –2 × 4 = –8. Again, Sn = n 2 [2a + (n – l)d] S20 = 20 2 [ 2(–8) + (20 – 1) 4] = 10 [ –16 +76] = 10 × 60 = 600. exercise 8.2 1. (a) Write the formula to calculate the sum of n terms of an A.S. in terms ‘a’, ‘n’ and ‘d’. (b) Write the relation among Sn, a, l and n. 2. (a) Find the first term, common difference and sum of 12 terms of the series 3 + 7 + 11 + 15 + ............ (b) Find the sum to 10 terms of an A.S. whose first term is 8 and the common difference is – 40. (c) Find the sum of 12 terms of an Arithmetic Sequence whose first 4 terms are 7,12,17 and 22. 3. (a) Find the sum of the series 24 + 28+ 32 + ..................... + 52. (b) Find the sum of the series 50 + 45 + 40 + ....................+ 20. (c) Find the sum of the series 5 + 8 + 11 + 14 + ..................... + 41 4. (a) Find the sum of the arithmetic series 50 + 45 + 40 + .............. to 5 terms. (b) Find the sum of the series 210 + 200 + 190 +.............. to 6 terms. 5. (a) Find the first term of an arithmetic progressive whose common difference is 3 and the sum of its first 20 terms is 670. (b) The sum of first 6 terms of an arithmetic sequence is –132. If its common difference is –5, find the first term (c) The first term of an Arithmetic series is 2 and the sum of its first 20 terms is 420, find the common difference.
Oasis School Mathematics-10 149 (d) The sum of first 10 terms of an arithmetic series is –50, of its first term is 50, find its common difference. 6. (a) The 5th term and 10th term of an arithmetic series are 17 and 42 respectively, find the sum of first 12 terms. (b) The 6th and 9th term of an Arithmetic series be 30 and 45 respectively. Find the sum of first 20 terms. 7. (a) The 9th term of an A.S is 19 and the sum of first 12 terms is 168, find the sum of first 8 terms. (b) The sum of first 7 terms of an A.S. is 14 and the sum of first 11 terms is 66, find the sum of first 25 terms. (c) The 6th term of an A.S is four times the second term. The sum of first 24 terms is 1704. find the sum of first 48 terms. (d) The fourth term of an Arithmetic series is 3 times, the first term and the 8th term of the series is 24. Find the sum of first 10 terms. (e) In an A.S., the sum of first 10 terms is 520. If the 7th term is double the third term, calculate the sum of first 20 terms. 8. (a) How many terms of an arithmetic sequence 3,6, 9, 12, .......... must be taken such that their sum of 234. (b) How many terms of the series 2 + 6+ 10 + 14 + .......... must be taken such that their sum is 128. 9. (a) The sum of 3 terms of an arithmetic series is 27 and their product is 648. Find the numbers. (b) The sum of 3 terms of an arithmetic series is 30. The product of the first and the last term is 75. Find 3 terms. (c) The sum of 3 terms of an arithmetic series is 69. If the product of first 2 terms is 483, find the three terms. 10. (a) The first and the last term of an arithmetic series are 5 and 30 respectively. If the sum of all terms of the series is 275. Find the number of terms and common difference. (b) The first and the last term of an arithmetic series are 2 and 4 respectively. If the sum of all the terms is 301, find the number of terms and the common difference. 11. The salary of a teacher of a private school during the time of his/her appointment is Rs. 25000. (a) If his/her salary increases by Rs. 1500 in the interval of 6 months, what will be his/her salary after 3 years. (b) If he/she has deposited all the salary in the bank till 5 years, how much money is deposited in his bank, account?
150 Oasis School Mathematics-10 (c) If Rs. 14,52,000 is collected in his bank account, how long has he/she worked there? (d) If another teacher appointed in the same school with starting salary Rs. 30,000 with same agreement, what will be his monthly salary after 4 years? 12. A man took loan of Rs. 1, 98,000. He paid Rs. 2500 in first month; Every month he paid Rs. 500 more than the previous payment.. (a) In how many instalment will he complete his loan? (b) How much money will he pay in the last instalment? Answers 1. Consult your teacher. 2. (a) 300 (b) –100 (c) 141 3. (a) 304 (b) 245 (c) 299 4. (a) 260 (b) 1110 5. (a) 5 (b) 1 (c) 2 (d) –10 6. (a) 294 (b) 1050 7. (a) 80 (b) 500 (c) 6864 (d) 165 (e) 1840 8. (a) 12 (b) 8 9. (a) 6, 9, 12 (b) 5, 10, 15 (c) 21, 23, 25 10. (a) n = 10, d = 5 (b) x = 14, d = 3. 11. (a) Rs. 32,500 (b) Rs. 19,05,000 (c) 4 years (d) Rs. 40,500 12. (a) 24 (b) Rs. 14000 Project Work Collect three different problems related to the sum of Arithmetic series which are connected to our daily life. geometric Series review • Are the terms 2, 4, 8, 16,........... in Geometric progression? • If the terms are in Geometric progression then guess the next 3 terms of the above sequence. • Discuss in the class, how can we find the next terms? • What is in the ration of second and first term? Similarly, what is the ratio of third to the second term? • The given set of numbers are in Geometric progression. • The ratio of two consecutive terms in the given sequence is always same. • This value is called common ratio. • Next term can be obtained by multiplying the terms by common ratio.
Oasis School Mathematics-10 151 remember ! • In a geometric series, tn = arn–1 Where, r = t 2 t 1 = t 3 t 2 = t 4 t 3 , ...... . • The common ratio of geometric sequence can be obtained by dividing second term by first term or third term and second term and so on. • Next term of geometric sequence can be obtained by multiplying given term by the common ration. geometric mean Let ‘a’ and ‘b’ be any two number. ‘G’ be their geometric mean. Then, a, G, b are in G.P. By the definition, G a = b G or, G2 = ab or, G = ab Hence, G.M. = ab ‘n’ geometric means between two numbers Let ‘a’ and ‘b’ be any two numbers and m1 , m2 , m3 ,, ..........mn be ‘n’ geometric means between ’a’ and ‘b’. Then, a, m1 , m2 , m3 , .....mn, b are in G.P. This geometric sequence contains (n + 2) terms. We have, t n = ar N-1 [ N = total number of terms] t n+2 = ar n+2-1 b a = rn+2 ∴ r = ( b a ) 1 n + 1 Here, m1 = t2 = ar = a ( b a ) 1 n + 1 m2 = t3 = ar2 = a ( b a ) 2 n + 1 m3 = t4 = ar3 = a ( b a ) 3 n + 1 ................................. ................................. mn = tn+1 = arn = a ( b a ) n n + 1
152 Oasis School Mathematics-10 Worked Out examples example 1 Find the geometric mean between 1 and 64. Solution: Here, a = 1, b = 64 We have, G.M. = ab = 1 × 64 = 64 = 8. example 2 Insert 3 g.M.’s between 1 8 and 2. Solution: Here, Let, m1 , m2 , m3 , be 3 G.M.’s between 1 8 and 2. Then, 1 8 , m1 , m2 , m3 , 2 are in G.P. Here, a = 1 8 , tn = 2, n = 5 We have, t n = arn-1 2 = 1 8 r5 - 1 or, 16 = r4 or, r = 16 4 = 2. Now, m1 = t2 = ar = 1 8 × 2 = 1 4 m2 = t3 = ar2 = 1 8 (2)2 = 1 8 × 4 = 1 2 m3 = t4 = ar3 = 1 8 (2)3 = 1 8 × 8 = 1. example 3 There are 5 g.M.’s between ‘a’ and ‘b’. If the second mean the last mean are 2 and 16 respectively, find the values of ‘a’ and ‘b’. Solution: Let, m1 , m2 , m3 , m4 , m5 be 5 G.M.’s between ‘a’ and ‘b’. Then, a, m1 , m2 , m3 , m4 , m5 b are in G.P.
Oasis School Mathematics-10 153 Now, Second mean (m2 ) = 2 t3 = 2 [ m2 = t3 ] ar2 = 2 or, a = 2 r2 ..............(i) Again, last mean (m5 ) = 16 t 6 = 16 [ m5 = t6 ] ar5 = 16 a = 16 r5 ............(ii) From equation (i) and (ii) 2 r2 = 16 r5 or, 2r5 = 16r2 or, r3 = 8 or, r = 2. Substituting the value of ‘r’ in (i) a = 2 r2 = 2 22 = 1 2 Now, b = t7 = ar6 = 1 2 (2)6 = 32 exercise 8.3 1. (a) Write the formula to calculate the G.M. between 2 numbers ‘a’ and ‘b’. (b) Find the geometric mean between two numbers 9 and 16. (c) If the geometric mean between ‘a’ and 25 is 20, find the value of ‘a’. 2. (a) Insert 3 G.M.’s between 9 and 1 9 . (b) Insert 4 G.M.’s between 2 3 and 162. (c) Insert 5 G.M.’s between 35 and 2240. (d) Insert 3 G.M.’s between 3 and 48. 3. (a) There are some geometric means between 1 2 and 16. If the third mean is 4, find the number of means.
154 Oasis School Mathematics-10 (b) There are some geometric means between 2 and 32. If the second mean is 8, find the number of means. 4. (a) Find the number of G.M.’s between 1 and 64 in which the ratio of first mean to the last mean is 1:16. (b) Find the number of G.M.’s inserted between 1 2 and 64 where the ratio of first mean to the last mean is 1:32. 5. (a) There are 5 G.M.’s between ‘a’ and ‘b’. If the second mean and the last mean are 2 and 16 respectively, find the values of a and b. (b) There are 5 G.M.’s between ‘a’ and ‘b’. If the first mean and last mean are 1 4 and 4, find the values of ‘a’ and ‘b’. 6. (a) Find the geometric mean between 4 and 256. (b) If there are 2 G.M.’s between 4 and 256, what are these means? (c) If there are 5 G.M.’s between these two numbers, find there means. 7. If (m + 2), (m + 8) and (m + 17) are in geometric sequence, (a) write the relation among 3 terms. (b) What is the value of ‘m’. (c) What is the common ratio of this sequence? (d) Write the next two terms of the sequence. 1. Consult your teacher. 2. (a) 1 3 , 1, 3, or –1 3 , 1, -3 (b) 2, 6, 18, 54 (c) 70, 140, 280, 560, 1120 or, -70, 140, -280, 560, -1120 (d) 6, 12, 24 3. (a) 4 (b) 3 4. (a) 5 (b) 6 5. (a) 1 2 , 32 (b) 1 8 , 8. 6. (a) 32 (b) 16, 64 (c) 4, 8, 16, 32, 64, 128 7. (b) 10 (c) 1 3 (d) 81 2 , 243 4 Answer The Sum of n terms of a geometric series Let ‘a’ be the first term of a Geometric series and ‘r’ be its common ratio. If l be the last term of the given geometric series. Then, Sum of n terms of a Geometric series Sn = a + ar + ar² + ar³ + .................... + l r + l ............ (i) Multiplying equation (i) by r, Then rSn = ar + ar² + ar³ + ar4 + .................... + l + lr ............ (ii) Subtracting equation (i) and (ii) rSn – Sn = lr –a
Oasis School Mathematics-10 155 Sn (r – 1) = lr – a Sn = lr – a r – 1 Hence, the sum of all the terms having first term ‘a’ and the last term ’l’ is Sn = lr – a r – 1 ............ (iii) We know that tn = arn–1 since last term (l) and nth term (tn) both are same; Then Sn = alrn–1. r – a r – 1 Or, Sn = arn – a r – 1 Sn = a(rn – 1) r – 1 Worked Out examples example 1 Find the sum of the series 2 – 4 + 8 – 16 + .... to 8 terms. Solution: Given series, 2 – 4 + 8 – 16 + .................. Here, first term (a) = 2 common ratio(r) = –4 2 = –2. number of terms (n) = 8. we have, Sn = a(rn – 1) r – 1 = 2 [(–2)8 – 1] –2 – 1 = 2 [256 – 1] –3 remember ! • Next term of a geometric series can be obtained by multiplying the given term by common ratio. • Previous term of a geometric series can be obtained by dividing the given term by common ratio. • nth term tn = arn–1 • Sum of the first n terms Sn = lr – a r – 1 Or, Sn = a(rn – 1) r – 1
156 Oasis School Mathematics-10 = 2 × 255 –3 = –2 × 85 = – 170. example 2 Find the sum of the series 2 + 6 + 18 + 54 + ...... + 4374 Solution: Given series, 2 + 6 + 18 + 54 + .......... + 4374 Here, first term (a) = 2. common ratio (r) = 6 2 = 3. last tern (l) = 4374. We have, Sum of the terms of geometric series, Sn = lr – a r – 1 = 4374 × 3 – 2 3 –1 = 13122 – 2 2 = 13120 2 = 6560. example 3 Find the number of terms and common ratio of the G.P whose first term is 7 last term is 189 and the sum the series is 280. Solution: Given, first term (a) = 7. last term (l) = 189 sum of the series (Sn) = 280. we have, Sn = lr – a r – 1 or, 280 = 189r – 7 r – 1 or, 280r – 280 = 189r – 7 or, 280r – 189r =280 – 7 or, 91r = 273 or, r = 273 91
Oasis School Mathematics-10 157 or, r = 3. Again, Sum of the first n terms. Sn = a(rn – 1) r – 1 or 280 = 7(3n –1) 3 – 1 or, 280 = 7(3n–1) 2 or, 560 = 7(3n –1) or, 560 7 = 3n – 1 or, 80 = 3n – 1 or, 80 + 1 = 3n or, 81 = 3n or, 34 = 3n or, n = 4. Hence, in common ratio(r) = 3 and the number of terms (n) = 4. example 4 Find the sum of 10 terms of a geometric progression whose 3rd term and 7th term are 4 and 1 4 respectively. Solution: Given, third term (t3 ) = 4. seventh term (t7) = 1 4 We have, t n = arn–1 t 3 = ar3–1 or, 4 = ar² or, a = 4 r² ......................(i) Again, t 7 = ar7–1 or, 1 4 = ar6 or, a = 1 4r6 ................... (ii) From equations (i) and (ii) 4 r² = 1 4r6 or, 16r6 = r² or, 16r4 =1
158 Oasis School Mathematics-10 or, r4 = 1 16 or, r4 = ( 1 2 ) 4 ∴ r = 1 2 . Substituting the value of r in (i) a = 4 r² a = 4 ( 1 2 )² or, a = 4 1 4 ∴ a = 16. Again, Sn = a(rn – 1) r – 1 S10 = 16[ 1 1024 – 1] – 1 2 = –16 [1 –1024 1024 × 2] = –16 [ –1023 1024 × 2] = 16 (1023 512 ) = 1023 32 . example 5 In a geometric series, the sumo of first 3 terms is 7 4 and the sum of first six terms is 63 32 , find the sum of first 10 terms. Solution: Given Sn = 7 4 when, n = 3. Sn = 63 32 when, n =6. Sn = ? when n = 10. We have, Sn = a(rn – 1) r – 1 7 4 = a(r3 – 1) r – 1 ............. (i) Again, S6 = a(r6 – 1) r – 1
Oasis School Mathematics-10 159 63 32 = a(r6 – 1) r – 1 ....................(ii) Dividing (ii) by (i) 63 32 7 4 = a(r6 – 1) r – 1 a(r3 – 1) r – 1 or, 63 32 × 7 4 = a(r6 – 1) r – 1 × r – 1 a(r3 – 1) or, 9 8 = a (r³ + 1) (r³ –1) a (r³ – 1) or, 9 8 = r³ + 1 or, 9 8 – 1 = r³ or, 1 8 = r³ or, ( 1 2 )³ = r³ ∴ r = 1 2 . Substituting the value of r in (i) 7 4 = a[( 1 2 )³ – 1] 1 2 – 1 or, 7 4 = a[ 1 8 – 1] 1 2 – 1 or, 7 4 = a(– 7 8 ) – 1 2 or, 7 4 = a × 7 8 ×2 or, 7 4 = a × 14 8 or, a = 8 × 7 4 × 14 or, a = 1. Again, Sn = a(rn – 1) r – 1 S10 = 1[( 1 2 ) 10 – 1)] 1 2 – 1 = 1 1024 – 1 – 1 2 = –1023 1024 – 1 2 = 1023 1024 × 2 = 1023 512 .
160 Oasis School Mathematics-10 example 6 In a G.P, the product of 3 numbers is 1000 and their sum is 35. Find the numbers. Solution: Let, three numbers in G.P be r a , a and ar. From the given condition, a r × a × ar = 1000 a³ = 1000 a³ = 10³ ∴ a = 10 Hence, the numbers is G.P be 10 r , 10 and 10r. Again, 10 r + 10 + 10r = 25 10 r + 10r = 25 or, 10 + 10r² r = 25 or, 10r² + 10 = 25r or, 10r² – 25r +10 = 0 or, 5(2r² – 5r + 2) = 0 or, 2r² + 5r + 2= 0 or, 2r² – (4 +1)r + 2 = 0 or, 2r² – 4r – r +2 = 0 or, 2r (r–2) –1 (r –2) = 0 or, (r – 2) (2r – 1) = 0. Either, r –2 = 0 ⇒ r = 2. or, 2r – 1 = 0 ⇒ 2r =1 or, r = 1 2 when, r = 2 Three terms are 10 r , 10 and 10r = 10 2 ,10 and 10 × 2 = 5, 10 and 20. when, r = 1 2 Three terms are 10 r , 10 and 10r = 10 1 2 , 10 and 10 × 1 2 = 20, 10 and 5.
Oasis School Mathematics-10 161 Example 7 In a g.P 6th term is 16 times the second terms and the sum of first seven terms is 127 2 . Find the positive common ratio and the first term. Solution: Given, t6 =16t2 and S7 = 127 2 . We have, t 6 = 16t2 . ar6–1 = 16 × ar2–1 ar5 = 16 × ar r5 = 16r r4 = 16 r4 = 24 ∴ r = 2. Again, Sn = a(rn – 1) r – 1 S7 = a(27 – 1) r – 1 or, 127 4 = a (128 –1) or, 127 4 = 127a a = 4 Hence, common ratio (r) =2, first term (a)= 4. example 8 There are 10 pencils in first row. 20 pencils in the second row and 40 pencils in the third row and so on. (a) If there are 6 rows, how many pencils are there altogether? (b) If there are 10,230 pencils, how many rows are required? (c) How many more pencils are there in 6th row then in 4th row ? Solution: Here, First term (a) = 10 Common ratio (r) = 20 10 = 2. (a) If there are 6 rows, n = 6. Now, Sn = a (rn – 1) r – 1 or, Sn = 10 (26 – 1) 2 – 1 = 10 (64 – 1) = 630.
162 Oasis School Mathematics-10 (b) Given, Sn = 10, 230, n = ?, r = 2, a = 10. We have, Sn = a (rn – 1) r – 1 or, 10, 230 = 10 (2n – 1) 2 – 1 or, 10,230 = 10 (2n– 1) or, 1023 = 2n – 1 or, 2n = 1024 or, 2n = 210 ∴ n = 10 Hence, 10 rows are required for 10,230pencils. (c) Now, tn = arn–1 or, t n = 10 × 26– 4 = 10 × 25 = 320 Again,t4 = 10 × 24–1 = 10 × 23 = 80. Hence, (320 – 80) = 240 more pencils are there in 6th row than in 4th row. exercise 8.4 1. (a) If ‘a’ be the first term. r be the common ratio and Sn be the sum of 2 terms of the geometric series, write the relation among ‘a’, ‘r’ Sn and n. (b) Write the relation among Sn, a, l and r 2. Find the sum of the following series. (a) 1 + 3 + + 9 ................to 8 terms. (b) 2 + 6 + 18 +.................to 6 terms. (c) 1 16 + 1 8 + 1 4 + ....................to 8 terms. (d) –8+ 4 – 2 + 1 + ................... to 10 terms. (e) 2 + 2 + 2 2 + ..................... + 8 (f) 81 + 27 + 9 + ......................... + 1 4 (g) 1 + 2 + 4 + 8....................+ 64 3. Find the value of: (a) ∑ 5 n=1 2n (b) ∑ 6 n=1 (-3)n (c) ∑ 7 n=1 (1 2 )n 4. (a) If the first term of a geometric series is 2 and the common ratio is 3. Find the sum of first 8 terms.
Oasis School Mathematics-10 163 (b) Find the sum of first 12 terms of a geometric series whose first term is 64 and the common ratio is 1 2 . (c) The first term,common ratio and the last term of the geometric series be 27,3 and 1 3 respectively. Find the sum of the series. 5. (a) Find the number of terms and the sum of the series 3 + 6+12 +........................+ 768. (b) In the series 4 + 8 + 16 + ................... +4096, find the number of terms and the sum of the terms of the series 6. (a) How many terms of the series 3 – 6 + 12 – 24 + .................. must be taken so that sum may be –63? (b) If the sum of the geometric series having first term 1 and common ratio 2 is 63. Find the number of terms of the series. (c) In a geometric progression the common ratio is 3, and the last term is 189. Find the number of terms whose sum is 280. 7. (a) Find the sum of first 10 terms of a G.P whose 3rd and 7th terms are 4 and 1 4 respectively. (b) If the 3rd term and 6th term of a geometric series be 12 and 16 respectively. Find the sum of first 8 terms. 8. (a) In a G.P, the sum of first two terms is 6 and the sum of first 4 terms is 30. Find the sum of 7 terms of the progression. (b) In a geometric series the sum of first three terms is 1 and the sum of its first 6 terms is 28. Find the first term, common ratio and the sum of first 10 terms of the series. (c) In a geometric series, if the 5th term is 8 times the second term and the sum of first four terms is 60, find the positive common ratio and the first terms of the series. (d) In the 7th term of a geometric series is 27 times the 4th term and the sum of first 2 terms is 8. Find the sum of first 10 terms. 9. (a) In a geometric series containing 8 terms, the sum of first 2 terms is 6 and the sum of last two terms is 384. Find the common ratio. (b) The sum of first four terms of a G.P is 30 and that of last four terms is 960. If the first term and the last term are 2 and 512 respectively, find common ratio. 10. In a Nursery, there are 4 flowers in a first bed, 8 flowers in second bed and 16 flowers in third bed and so on. (a) If there are 7 beds in the Nursery, how many flowers are there altogether? (b) If there are 1020 flowers in the Nursery, how many beds are required? (c) How many more flowers are there in the 5th bed than in the 3rd bed?
164 Oasis School Mathematics-10 11. A sum of Rs. 19,375 is to be distributed among top 5 students of the class as a scholarship. If the scholarship get by second ranked student is half of the scholarship get by the first student and so on. (a) If the 5th student get the scholarship of Rs. 625, what is the scholarship get by the first student? (b) If the scholarship in the same format is distributed in another school, such that first student gets Rs. 20,000. How much budget is required for if the scholarship is given for 5 students. 12. Aadhya’s father deposited Rs. 5000 in the bank is her 5th birthday. On her 6th birthday he deposited two times the amount deposited in the 5th birthday. If he deposited the money in this way, (a) how much money will he deposit in her 10th birthday? (b) how much money is collected in the bank at her 10th birthday? (Neglect the interest) (c) on which birthday he will deposit Rs. 40,000? Project Work Collect some problems related to the sum of geometric series which is related to our daily life. 1. Consult your teacher 2. (a) 3280 (b) 728 (c) 255 6 (d) 2728 (e) 14+7 2 (f) 364 3 g) 127 3. (a) 62 (b) 564 (c) 127 128 4. (a) 6560 (b) 4095 32 (c) 121 3 5. (a) 3, 1533 (b) 11, 1188 6. (a) 6 (b) 6 (c) 4 7. (a) 1023 512 (b) 765 8. (a) 254 (b) 19049 26 (c) 2, 4 (d) 59048 9. (a) 1023 (b) 2 10. (a) 508 (b) 8 (c) 48 11. (a) Rs. 10,000 (b) Rs. 38,750 12. (a) Rs. 1,60,000 (b) Rs. 3, 15,000 (c) 8th Answers
Oasis School Mathematics-10 165 9.1 Warm-up Activities • What are the degree of the following equations? x2 – 9 = 0, x2 –5x+6=0, x2 –29=7, (x+1) (x+2) = 0 • What are these equations called? • Guess two values of x which satisfy the given equations? • Compare the above equations with ax2 +bx+c=0 and find the values a, b and c. • Factorise the above equations. In the above given examples, the degree of variable x is 2. So, these equations are quadratic equations. Hence, the quadratic equation is defined as which contains the squares of the unknown quantities. The general form of quadratic equation is ax2 + bx + c = 0 where a, b, c are real numbers, and a ≠ 0. (a) Solution of quadratic equation by the method of factorisation The general form of quadratic equation is, ax2 + bx + c = 0, where a, b and c are real numbers. The method for solving such equation is based on the following property. If a × b = 0, then either a = 0 or b = 0. This principle is called zero – factor property. The steps for solving a quadratic equation by the method of factorisation are as follows: • Express the given equation in the general form ax2 + bx + c = 0 (simplify if necessary). • Factorise the left side. • Equate either of the factor equal to zero. • Solve them and find two values of x. Unit 9 Quadratic equation
166 Oasis School Mathematics-10 Worked Out examples example 1 Solve : 4x² = 25 Solution: Here, 4x2 = 25 or, 4x2 – 25 = 0 or, (2x)2 – (5)2 = 0 or, (2x + 5) (2x – 5) = 0 Either, 2x + 5 = 0 .......... (i) or, 2x – 5 = 0 ............(ii) From equation (i), we get, 2x + 5 = 0 or, 2x = – 5 or, x = – 5 2 example 2 Solve : 6x² + x = 12 Solution: Here, 6x2 + x = 12 or, 6x2 + x – 12 = 0 or, 6x2 + (9 – 8) x – 12 = 0 or, 6x2 + 9x – 8x – 12 = 0 or, 3x (2x + 3) – 4(2x + 3) = 0 or, (2x + 3) (3x – 4) = 0 Either, 2x + 3 = 0 .......... (i) or, 3x – 4 = 0 .......... (ii) From equation (i), we get, 2x + 3 = 0 or, 2x = – 3 or, x = – 3 2 Similarly, from equation (ii) we get, 2x – 5 = 0 or, 2x = 5 or, x = 5 2 ∴ x = ± 5 2 Similarly, From equation (ii), we get, 3x – 4 = 0 or, 3x = 4 or, x = 4 3 ∴ x = – 3 2 or, 4 3
Oasis School Mathematics-10 167 example: 3 Solve : 3 x + 1 + 1 x – 1 = 5 4 Solution: Here, 3 x + 1 + 1 x – 1 = 5 4 or, 3(x – 1) + (x + 1) (x + 1) (x – 1) = 5 4 or, 3x – 3 + x + 1 x2 – 1 = 5 4 or, 4x – 2 x2 – 1 = 5 4 or, 5x2 – 5 = 16x – 8 or, 5x2 – 16x + 3 = 0 5x2 – 15x – x + 3 = 0 or, 5x (x – 3) – 1 (x – 3) = 0 or, (x – 3) (5x – 1) = 0 Either, x – 3 = 0 .......... (i) or, 5x – 1 = 0 .......... (ii) From equation (i), we get, Similarly, from equation (ii), we get, x – 3 = 0 5x – 1 = 0 or, x = 3 or, 5x = 1 ∴ x = 3 and 1 5 or, x = 1 5 (b) Solution of quadratic equation by completing the perfect square The method of completing the square can be applied to all quadratic equations. The process of this method is given below: • Transpose the constant term to the right–hand side of the equations: • Divide both sides of the equations by coefficient of x2 . • Make the left side of the equation a perfect square. • Take the square root of each side. • Solve the resulting simple equations. example 4 Solve : x² – 5x + 6 = 0 Solution: Here, x2 – 5x + 6 = 0 or, x2 – 5x = – 6
168 Oasis School Mathematics-10 or, x2 – 2.x. 5 2 + 5 2 2 = – 6 + 5 2 2 or, x – 5 2 2 = 25 4 – 6 or, x – 5 2 2 = 25 – 24 4 or, x – 5 2 2 = 1 4 or, x – 5 2 2 = ± 1 2 2 or, x – 5 2 = ± 1 2 Taking positive sign, Taking negative sign, x = 5 2 + 1 2 x = 5 2 – 1 2 or, x = 6 2 or, x = 4 2 ∴ x = 3 ∴ x = 2 ∴ x = 3 or 2 example 5 Solve : 3x² – x – 2 = 0 Solution: Here, 3x2 – x – 2 = 0 Dividing both sides by 3 3x2 3 – x 3 – 2 3 = 0 or, x2 – x 3 – 2 3 = 0 or, x2 – 2.x. 1 6 + 1 6 2 – 2 3 = 1 6 2 or, x – 1 6 2 = 1 36 + 2 3 or, x – 1 6 2 = 25 36 or, x – 1 6 2 = ± 5 6 2 or, x – 1 6 = ± 5 6
Oasis School Mathematics-10 169 ∴ x = ± 5 6 + 1 6 Taking positive sign Taking negative sign or, x = 5 6 + 1 6 or, x = 1 6 – 5 6 = 6 6 = – 2 3 = 1 ∴x = 1 or – 2 3 (c) Solution of quadratic equation by using formula Every quadratic equation can be expressed in the form of ax2 + bx + c = 0. To find the values of x that satisfy the equation, we use the method of completing the square, ax2 + bx + c = 0 or, ax2 + bx = – c or, x² + b a x = – c a [Dividing both sides by a.] or, x2 + 2. x . b 2a + b 2a 2 = – c a + b 2a 2 [Adding b 2a 2 to both sides] or, x + b 2a 2 = b2 4a2 – c a or, x + b 2a 2 = (± b2 – 4ac 4a2 ) 2 or, x + b 2a = ± b2 – 4ac 2a ∴ x = ± b2 – 4ac 2a – b 2a or, x = – b ± b2 – 4ac 2a From the above example, we have two solutions for unknown quantity x which are given as: – b + b2 – 4ac 2a and – b - b2 – 4ac 2a , where a, b and c are real number and a ≠ 0. This general formula is used in solving a quadratic equation. It can be consequently applied when the left side cannot be factorised.
170 Oasis School Mathematics-10 example 6 Solve: x² + 2x – 15 = 0. Solution: Here, given quadratic equation is x2 + 2x – 15 = 0 comparing this equation with ax2 + bx + c = 0 a = 1, b = 2 and c = –15 We have, x = – b ± b2 – 4ac 2a x = – 2 ± 22 – 4 × 1 × (– 15) 2 × 1 = – 2 ± 4 + 60 2 = – 2 ± 64 2 = – 2 ± 8 2 Now, taking +ve sign, x = – 2 + 8 2 = 6 2 = 3 Similarly, taking –ve sign, x = – 2 - 8 2 = – 10 2 = – 5 ∴ x = 3 and – 5 Example 7 Solve: 4x² – 3x = 6 Solution: Here, 4x2 – 3x = 6 4x2 – 3x – 6 = 0 Now, comparing the given equation with ax2 + bx + c = 0, we get a = 4, b = – 3 and c = – 6 We have, x = – b ± b2 – 4ac 2a ∴ x = – (– 3) ± (–3)2 – 4 × 4 × – 6 2 × 4 = 3 ± 9 + 96 8 = 3 ± 105 8 Now, taking + ve sign, we get, x = 3 + 105 8
Oasis School Mathematics-10 171 Similarly, taking –ve sign, we get, x = 3 - 105 8 ∴ x = 3 + 105 8 or, x = 3 - 105 8 example: 8 Solve : x x – 1 + x – 1 x = 2 1 2 Solution: Here, the given equation is, x x – 1 + x – 1 x = 2 1 2 or, x2 + (x – 1)2 x(x – 1) = 5 2 or, x2 + x2 – 2x + 1 x2 – x = 5 2 or, 2 (2x2 – 2x + 1) = 5 (x2 – x) or, 4x2 – 4x + 2 = 5x2 – 5x or, 4x2 – 4x + 2 – 5x2 + 5x = 0 or, – x2 + x + 2 = 0 or, –(x2 – x – 2) = 0 or, x2 – x – 2 = – 0 Now, comparing this equation with ax2 + bx + c = 0, we get, a = 1, b = – 1 and c = – 2 Using the formula, x = – b ± b2 – 4ac 2a ∴ x = – (– 1) ± (–1)2 – 4 × 1 × – 2 2 × 1 = 1 ± 1 + 8 2 = 1 ± 3 2 Now, taking + ve sign we get, x = 1 + 3 2 = 4 2 = 2 Similarly, taking –ve sign we get, x = 1 - 3 2 = –2 2 = – 1 Hence, x = 2 and – 1.
172 Oasis School Mathematics-10 exercise 9.1 1. (a) If (x + 1) (x – 1) = 0, find the values of x. (b) What are the values of x in the quadratic equation ax2 + bx + c = 0. (c) If x (x–2) = 0, what are the values of x? 2. Solve: (a) (x – 1) (x + 2) = 0 (b) (x – 5) (2x + 3) = 0 (c) (3x – 1) (x + 4) = 0 (d) (5 – 2x) (3x – 7) = 0 (e) (5x – 1) (2x +7) = 0 3. Solve: (a) x2 – 2x = 0 (b) 4x2 + 3x = 0 (c) x2 – 7x = 0 (d) x2 – 9 = 0 (e) 9x2 – 16 = 0 (f) x2 = 4x (g) 5x2 = 125 4. Solve (by factorisation method): (a) x2 – 10x – 24 = 0 (b) x2 – 5x – 6 = 0 (c) 5x2 – 2x – 3 = 0 (d) 3y2 + 7y – 6 = 0 (e) (4x – 3) (8x + 5) = 13 (f) x2 + 50 25 = 7 – 2x2 + 10 15 (g) 5x + 7 x – 1 = 3x + 2 (h) x + 1 x – 1 – x – 1 x + 1 = 7 12 (i) x x + 1 + x + 1 x = 2 1 12 (j) 4 x – 1 – 5 x + 2 = 3 x (k) x x + 1 + x + 1 x = 13 6 (l) x + 3 x + 2 – 2x – 3 x – 1 = x – 3 2 – x (m) 2 + x 2 – x + 2 – x 2 + x = 17 4 5. Solve (by completing the square method): (a) x2 – 10x + 21 = 0 (b) x2 + 6x – 40 = 0 (c) x2 – 5x – 36 = 0 (d) x2 – 4x + 1 = 0 (e) 2x – 3 3x – 7 = x + 2 x + 3 (f) x + 5 5x – 2 + x – 2 3x – 8 (g) x + 2 x – 2 + 2x – 3 2(x – 1) = 23 6 (h) 4 x – 1 – 5 x + 2 = 3 x 6. Solve (by using formula): (a) x2 - 11x + 30 = 0 (b) 3x2 – 10x – 13 = 0 (c) 7x2 + 2 = 9x (d) (4x - 3)2 = 16x (e) 2x2 – 1 6 = 2 – 3x2 15 (f) x x + 1 + x + 1 x = 5 2 (g) 3 x – 1 + 2 x – 3 = 2 x
Oasis School Mathematics-10 173 7. Solve (using any method): (a) 6 x2 - 4x - 2 6 = 0 (b) (2x + 3) ( 3 - x) = 2x – 6 (c) 1 x – 1 + 2 x = 3 2 – x (d) x – 5 2 x – 4 = 4 – 10 x (e) 1 3x + 2 + 3 x + 6 = 1 x (f) x – 1 x – 2 + x – 3 x– 4 = 3 1 3 8. (a) Convert the equation x + 2 x – 2 – x – 2 x + 2 = 4 4 5 in the form of ax² + bx + c = 0. (b) Factorise the expression thus obtained. (c) What are the possible solution of this equation? 9. (a) Convert the equation 3 x – 1 + 2 x – 3 = 2 x in the form of ax² + bx + c = 0. (b) Convert the equation thus obtained into perfect square on both left and right side of the equation. (c) Find the possible solution of the equation. 10. (a) Convert the equation x x + 2 + x + 2 x = 4 in the form of ax² + bx + c = 0. (b) Write the formula to get the value of x in the quadratic equation ax² + bx + c = 0. (c) Using this formula, calculate the value of x obtained in the equation (a). Answers 1. Consult your teacher. 2. (a) 1, –2 (b) 5, – 3 2 (c) 1 3 , – 4 (d) 5 2 , 7 3 (e)1 5 , –7 2 3. (a) 0, 2 ( b ) 0, – 3 4 (c) 0, 7 (d) 3, –3 (e) 4 3 , -4 3 (f) 0, 4 (g) 5, – 5 4. (a) 12, –2 ( b ) 6, –1 (c) 1, – 3 5 (d) – 3, 2 3 (e) 1, – 7 8 (f) 5, –5 (g) 3, – 1 (h) 7, –1 7 (i) – 4, 3 (j) 3, –1 3 (k) 2, –3 (l) 0, 4 (m) ± 6 5 5. (a) 7, 3 (b) 4,– 10 (c) 9, –4 (d) 2 ± 3 (e) 5, –1 (f) 4, 11 2 (g) 4,10 11 (h) 3, – 1 2 6. (a) 5, 6 (b) 13 3 , – 1 (c) 1, 2 7 (d) 9 4 , 1 4 (e) ± 3 4 (f) 1, –2 (g) 2, – 1 7. (a) 6 , – 6 3 (b) 3, –5 2 (c) 4 3 , 1 2 (d) 5 2 , 16 3 (e) 2, – 6 7 (f) 5, 5 2 8. (c) 3, – 4 3 9. (c) 4, 0 10. (c) – 1 + 3 , –1 – 3 .
174 Oasis School Mathematics-10 9.3 Quadratic equations Discuss the following with your friend and draw a conclusion. Solve the given problem: x2 – 4 = 0. • What are the values of x? Do both the values satisfy the given equation or not? • What are the factors of x2 – 3x + 2? Solve the equation x2 – 3x + 2 = 0. • What are the solutions to the above equation? Do both values satisfy the given equation? • Solve the equation x2 – 4x + 2 = 0. • How do you factorise x2 – 4x + 2 ? • If it is not factorised, how do you get the values of x? Give some examples of quadratic equations. remember ! In the quadratic equation ax2 +bx+c=0 , a ± 0 x = –b ± b2 – 4ac 2a If the given equation is not factorised, use this formula to get the values of x. Some problems like the age of two people, area of rectangles, formation of two-digit numbers are related in such a way that their relation can be expressed in the form of an quadratic equation. These quadratic equations can be solved to get the required values. Every quadratic equation has two roots. Both roots may not satisfy the given equation. The root which doesn't satisfy the given condition is to be rejected. Worked Out examples example 1 If 4 is subtracted from half of the square of a natural number, the result is 14. Find the number. Solution: Let the required number be 'x'. From the given condition, 1 2 x² – 4 = 14 or, 1 2 x² = 14 + 4 or, 1 2 x² = 18 or, x² = 36
Oasis School Mathematics-10 175 or, x² = (± 6)² or, x = ± 6 Hence, the required natural number is 6. example 2 The sum of two numbers is 12 and the sum of their squares is 80. Find the numbers. Solution: Let the required numbers be 'x' and 'y'. From the given condition, x + y = 12 or, y = 12 – x................. (i) Again, from the given condition, x² + y² = 80 or, x² + (12 – x)² = 80 or, x² + 144 – 24x + x² = 80 or, 2x² – 24x + 144 – 80 = 0 or, 2x² – 24 x + 64 = 0 or, x² – 12x + 32 = 0 or, x² – 8x – 4x + 32 = 0 or, x (x – 8) – 4 (x – 8) = 0 or, (x – 8) (x – 4) = 0 Either, x – 8 = 0 or, x – 4 = 0 or, x = 8 or, x = 4 When, x = 8, from equation (i), y = 12 – x or, y = 12 – 8 or, y = 4 Again, when, x = 4 y = 12 – x or, y = 12 – 4 y = 8 ∴ The required numbers are 4 and 8 or 8 and 4 example 3 Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller by 164. Solution: Let the larger part = x then the smaller part = (16 – x) Now, from the given condition,
176 Oasis School Mathematics-10 2x2 – (16 – x)2 = 164 or, 2x2 – (256 – 32x + x2 ) = 164 or, x2 + 32x – 420 = 0 or, x2 + 42x – 10x – 420 = 0 or, x (x + 42) – 10 (x + 42) = 0 or, (x + 42) (x – 10) = 0 Either, (x + 42) = 0, or, x = – 42 (neglected) or, x – 10 = 0, or, x = 10 ∴ Larger part = 10 Smaller part = (16 – 10) = 6 example 4 The product of two consecutive odd numbers is 143. Find the numbers. Solution: Let the two consecutive odd numbers be x and x + 2. From the given condition, x(x + 2) = 143 x² + 2x = 143 x² + 2x – 143 = 0 x² + 13x – 11x – 143 = 0 or, x (x + 13) – 11 (x + 13) = 0 (x + 13) (x – 11) = 0 Either, x + 13 = 0 or, x = – 13 (neglected) or, x – 11 = 0 or, x = 11 ∴ Two numbers are x = 11 and x + 2 = 11 + 2 = 13 example 5 The present ages of the elder and younger brothers are 13 years and 7 years respectively. (a) What will be their ages after ‘t’ years? (b) If the product of their ages after t years is 280, form the equation quadratic in ‘t’. (c) Find the value of ‘t’ and find after how many years product of their ages will be 280? Solution: (a) Their ages afters ‘t’ years be (13 + 1) yrs and (7 + 1)yrs. (b) From the given condition, (13 + t) (7 + t) = 280 or, 91 + 13t + 7 t + t² = 280 or, t² + 20 t + 91 = 280
Oasis School Mathematics-10 177 or, t² + 20 t + 91 – 280 = 0 or, t² + 20t – 189 = 0 (c) Again, factorising this equation, or, t² + 27 t – 7t – 189 = 0 or, t(t + 27) – 7( t + 27) = 0 or, (t + 27) (t – 7) = 0 Either, t + 27 = 0 or, t = – 27 ( not possible) or, t – 7 = 0 or, t = 7 Hence, t = 7 Hence, after 7 years the product of their ages will be 280. example 6 The product of the ages of a father and his son is 180 years. (a) Convert this statement into equation. (b) After how many years, son will be as old an his father now?, what will be their ages at that time? (c) If the sum of their ages at that time will be 84, find their present ages. Solution: (a) Let the present age of the father and his son be x years and y years respectively. (b) From the given condition, xy = 180 .............(i) The son will be as old as his father now after (x – y) years. (c) x + (x - y) + y + (x – y) = 84 3x – y = 84 y = 3x – 84 ..........(ii) From (i) and (ii) xy = 180 or, x(3x – 84) = 180 or, 3x2 – 84x = 180 or, 3x2 – 84x –180 = 0 or, 3(x2 – 28x – 60) = 0 or, x2 – 28x – 60 = 0 or, x2 – (30 – 2) x – 60 = 0 or, x2 – 30x + 2x – 60 = 0 or, x(x – 30) + 2(x – 30) = 0 or, (x – 30) (x + 2) = 0 Either x – 30 = 0 ⇒ x = 30 or, x + 2 = 0 ⇒ x = –2 (not possible) From (i) xy = 180
178 Oasis School Mathematics-10 or, 30y = 180 or, y = 6 ∴ Father’s age = 30 years, Son’s age = 6 years Example 7 The product of the digits of a two-digit positive number is 21. If 36 is added to the number, the digits will interchange their places. (a) Convert above statements into equation. (b) Solve the above equations (c) Find the required number. Solution: (a) Let x be the digit in the place of tens and y be the digit in place of ones. Then the number is (10x + y). If the digits interchange their places, then the new number is (10y + x). (b) From the condition, xy = 21 ..................(i) Also, 10x + y + 36 = 10y + x or, 10x + y – 10y – x = –36 or, 9x – 9y = –36 or, 9 (x – y) = –36 or, x – y = –4 or, y = x+4 ..............(ii) (c) From (i) and (ii) or, xy = 21 or, x(x + 4) = 21 or, x2 + 4x = 21 or, x2 + 4x – 21 = 0 or, x2 + 7x – 3x – 21 = 0 or, x(x + 7) – 3(x + 7) = 0 or, (x + 7) (x – 3) = 0 Either x + 7 = 0 or, x = –7 (not possible) or, x – 3 = 0 or, x = 3 From (i) xy = 21 3xy = 21 y = 7 ∴The required number = 10x + y = 37. example 8 The perimeter of a rectangle is 72 m and the area is 320 m². (a) Convert the above statements into equations. (b) Solve these two equations to get the length and breadth of thee rectangle.
Oasis School Mathematics-10 179 Solution: (a) Let the length and breadth of the rectangle be x m and y m respectively. Then, perimeter of the rectangle = 2(x + y) ∴ 2(x + y) = 72 or, x + y = 36 or, y = ( 36 – x) ………….. (i) Since, the area of rectangle is 320 m2 xy = 320 (b) Solving equation (i) and (ii) or, x(36 – x) = 320 or, 36x – x2 = 320 or, –x2 + 36x – 320 = 0 or, –(x2 – 36x + 320) = 0 or, x2 – 36x + 320 = 0 or, x2 – 20x – 16x + 320 = 0 or, x(x – 20) – 16(x – 20) = 0 or, (x – 20) (x – 16) = 0 Either x – 20 = 0 or x = 20 or, x – 16 = 0 or, x = 16 when x = 20 y = 36 – 20 = 16 when, x = 16 y = 36 – 16 = 20. Taking the longer part as length, Length of the rectangle = 20 m and the breath of the rectangle = 16 m. Example 9 Students of class X planned a picnic with a fixed budget (a) If 10 more students were there expense per student is reduced by Rs. 100. Covert this statement into the equation. (b) If 15 students did not join the picnic, expense per students increased by Rs. 300. (c) Solve above two equations to get the number of students in the class and expense per student if all students join the picnic. (d) Find the total budget of the picnic. Solution: (a) Let, the number of students in the class = x and expense per student if all of them joined the picnic = Rs. y. Total budget = xy.
180 Oasis School Mathematics-10 In the first condition, If the number students = (x + 10), expense per students = (y – 100) Now, (x + 10) (y – 100) = xy xy – 100x + 10y – 1000 = xy or, – 100x + 10y = 1000 or, –10 (10x – y) = 1000 or, 10x – y = – 100 or, y = 10x + 100 .................... (i) (b) If 15 students did not join the picnic, number of students = (x – 15) expense per student = (y + 300) Now, (x – 15) (y + 300) = xy xy + 300x – 15y – 4500 = xy 300x –15y = 4500 15(20x – y) = 4500 20x – y = 300 ...................... (ii) (c) Substituting the value of y from (i) in (ii) 20x – y = 300 20x – (10x + 100) = 300 20x – 10x – 100 = 300 10x = 300 + 100 x = 40. Substituting the value of x in eqn (i) y = 10x + 100 = 10 × 40 + 100 = 500 Hence, if all 40 students joined the picnic each students has to pay Rs. 500. (d) Total budget of the picnic = xy = 40 × 500 = 20,000 exercise 9.3 1. (a) If ‘x’ be a natural number, what are the three consecutive natural numbers that are greater than x?
Oasis School Mathematics-10 181 (b) ‘x’ is an odd number. What are the three consecutive odd numbers greater than x. (c) ‘x’ is an even number. What are the three consecutive even numbers less than x? (d) ‘x’ is a number which is a multiple of 5. What are the three consecutive number less than x which are a multiple of 5. (e) The present ages of a father and his son are ‘x’ and ‘y’ years respectively. How many years before was the father as old as his son? (f) The sum of a number and its reciprocal is 13 6 , make the equation of this statement. 2. (a) If 6 is added to the square of a number, the result is 31. Find the numbers. (b) If 3 is subtracted from the square of a number, the result is 6. Find the numbers. (c) If 10 is added to two times the square of a positive number, the result is 60. Find the number. (d) If 5 times a number is subtracted from the square of a number, the result is 6. Find the numbers. (e) If a number is subtracted from its square, the result is 2. Find the numbers. (f) If 4 times a natural number is equal to the square of a number, find the numbers. (g) If 8 times a natural number is equal to the square of the number, find the number. (h) The sum of a number and its reciprocal is 2 1 2 . Find the number. (i) Find the number such that the sum of the number and 48 times its reciprocal is equal to 19. 3. (a) Two natural numbers are in the ratio 3:5. If their product is 375, find the numbers. (b) Two natural numbers are in the ratio 3:2. If the difference of their square is 45, find the numbers. (c) Two natural numbers are in the ratio 1:2 of the sum of their squares which is 125, find the numbers. 4. (a) Find the two numbers whose sum is 5 and the product is 6. (b) Divide 57 into two parts whose product is 350. (c) Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164. (d) The product of two consecutive positive integers is 56. Find the numbers. (e) The product of two consecutive positive even numbers is 288. Find the numbers. 5. (a) If the difference between the squares of two consecutive natural numbers is 5, find the numbers. (b) Find the two consecutive positive even integers, the sum of whose squares is 340. 6. (a) The present ages of a boy and his younger brother are 13 years and 7 years respectively. How many years ago was the product of their ages 55 ? (b) The present ages of a father and his son are 36 years and 12 years respectively. In how many years will the product of their ages be 640? (c) The difference of the present ages of two sisters is 5 years. After 4 years if the product of their ages becomes 104, find their ages. (d) One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son's age. Find their present ages.
182 Oasis School Mathematics-10 (e) The ratio of the present ages of two bothers is 2:1. Three years hence, the ratio of the square of their ages will be 9:4. Find their present ages. (f) The difference of the present ages of a sister and her younger brother is 6 years. The product of their ages is equal to 4 times the sum of their ages. Find the present ages of the sister and her brother. (g) The product of the present ages of a mother and her daughter is 100. 10 years later, the mother's age will be twice that of the daughter. Find their present ages. (h) The ages of Rupsi and Rushma are in the ratio of 5:7. Eight years ago, their ages were in the ratio of 7:13. Find their present ages. (i) The product of the present ages of a father and his son is 96 years. When the son becomes as old as his father is now, the sum of their ages will be 68 years. Find their present ages. (j) The age of a mother is 5 years less than twice the age of her daughter. When the mother was as old a her daughter is now, the product of their ages was 125. Find their present ages. 7. (a) A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the numbers. (b) A number between 10 to 100 is four times the sum and three times the product of its digits. Find the number. (c) A two–digit number is such that the product of the digits is 16. When 54 is subtracted from the number, the positions of the digits interchange. Find the number. (d) A number of two-digits is equal to 4 times the sum of the digits. If the product of the digits is 8, find the number. (e) The product of the digits of a two-digit number is 18. The number formed by interchanging the digits of the numbers will be 27 more than the original number. Find the original number. 8. (a) The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is 2 9 10 . Find the fraction. (b) The numerator of a fraction is 1 less than the denominator. If the sum of the fraction and its reciprocal is 13 6 , find the fraction. 9. (a) A person on tour has Rs. 360 for his daily expenses. If he extends his tour programme by 4 days, he must cut down on his daily expenses by Rs. 3 per day. Find the number of days of his tour programme. (b) Some boys planned a picnic with a total budget of Rs. 4,200. But 5 boys cannot attend the picnic and the cost of picnic increased by Rs. 70 for each boy. How many boys joined the picnic? 10. (a) The lengths of the sides of a right-angled triangle are (x + 2) cm, 4x cm and (4x + 1) cm. Find the length of the sides of the triangle. (b) The hypotenuse of a right angled-triangle is 6 meters more than twice the shortest side. If the third side is 2 meters less than the hypotenuse, find the sides of the triangle. (c) The hypotenuse of a right-angled triangle is 20 m. If the difference between the lengths of the other sides is 4 meters, find the perimeter of the triangle.
Oasis School Mathematics-10 183 11. (a) If the perimeter of a rectangular plot is 104 meters and its area is 640 m², find its length and breadth. (b) If the rectangular room has a perimeter 60 m and area 221 m², find the length and breadth of the room. 12. (a) If I had walked 1 km per hour faster, it would have taken 100 minutes less to walk a distance of 20 km. Find the speed of my walking. (b) A train takes 3 hours less for a journey of 360 km if its speed is increased by 10 km/hr from its usual speed. What is the usual speed? (c) A bus takes 3 hours less time than another bus to complete a journey of 360 km. If the speed of the first bus is 20 km/hr more than that of the second bus, find their speed. 13. Father’s age was four time the age of his son on 2050 B.S. In 2068 B.S., product of their ages in 1458. (a) Assuming the ages of father and son x years and y years in 2050 B.S. make two equations of above two statements. (b) Solve the above two equation to get the value of ‘x’ and ‘y’. (c) Find the birth year of father and son. (d) What will be the ratio of their ages in 2086 B.S.? 14. At the age of 25 Sarmila gave a birth to her son. If the product of their present ages is 600, (a) Make the equation from the given condition. (b) Solve the above equation to get the ages of mother and her son. (c) After how many years, product of their ages will be 900? 15. The age of father is the square of the age of his son. (a) Convert this statement into the equation. (b) If the 5 times the age of his son is added to the father’s age, the result is 66 years. Convert this statement into the equation. (c) Solve the equation obtained in (a) and (b) to get the ages of father and his son. (d) Before how many years, product of their ages is 64 years? 16. If the age of father is x yrs and the age of son is y years, (a) After how many years the age of son will be as old as his father now? (b) The product of their ages is 180. When the son will be as old as his father now, sum of their ages will be 84. Convert above statement into equation. (c) Solve the equations obtained in (b) to get the age of father ans son. (d) After how many years, product of their ages will be 340 ? 17. In a two - digit number, the digit in the ones place is 1 more than square of the digit of ten’s place. (a) Convert this statement into the equation. (b) The position of the digits reversed if 2 is added to the two times the original number. Convert this statement into equation.
184 Oasis School Mathematics-10 (c) Solve the above the equations to get the digits in tens and ones place. (d) Find the number. (e) What is the product of the digits in the number? 18. The total cost of 5 kg rice and 6 kg sugar is Rs. 940. If the rate of rice increases by 20% and the rate of sugar decreases by 10%, the total cost of 4kg rice and 3 kg sugar is Rs. 627. (a) Convert the above two statement into equation. (b) Solve above two equation to get the price per kg rice and price per kg sugar? (c) By what percent cost of per kg rice is more or less than cost per kg sugar. 19. A person in tour has total budget of Rs. 50,000. He is planning his tour for x days and expense per day Rs. y. (a) Convert the above statement into the equation. (b) If he wants to extend his hour tour programme by 5 days he must cut down his daily expense by Rs. 500 per day, convert this statement into equation. (c) Solve the equation obtained in (a) and (b) to get the values of x and y. (d) If he reduces his daily expense by Rs. 1250 per day, for how many days his tour can be extended? Answers 1. Consult your teacher 2. (a) 5,-5 (b) 3, -3 (c) 5 (d) 6, -1 (e) 2, -1 (f) 4 (g) 8 (h) 2, 1/2 (i) 16,3 3. (a) 15, 25 (b) 9, 6 (c) 5, 10 4. (a) 3, 2 (b) 50, 7 (c) 10, 6 (d) 7, 8 (e) 16, 18 5. (a) 2, 3 (b) 12, 14 6. (a) 2 years (b) 4 years (c) 9 years, 4 years (d) 49 years, 7 years (e) 6 years, 3 years (f) 12 years, 6 years (g) 20 years, 5 years (h) 15 years, 21 years (i) 24 years, 4 years (j) 45 years, 25 years 7. (a) 92 (b) 12 (c) 82 (d) 24 (e) 36 8. (a) 2 5 (b) 2 3 9. (a) 20 days (b) 15 10. (a) 5 cm, 12 cm, 13 cm (b) 10 m, 24 m, and 26 m, (c) 48 m 11. (a) 32 m, 20 m (b) 17m, 13m 12. (a) 3 km/hr. (b) 30 km/hr. (c) 60 km/hr, 40 km/hr. 13. (b) 9 yrs, 36 yes (c) 2014 B.S and 2041 B.S (d) 5 : 8 14. (b) 40 yrs, 15 yrs, (c) 5 yr 15. (c) father 36 yrs, son: 6 yrs (d) 2 yrs 16. (c) 30 years, 6 years (d) 4 yrs 17. (c) Digit in tens place = 2, Digit in ones place = 5 (d) 25 (e) 10 18. (b) Rice : Rs. 80 per kg, Sugar Rs. 90 per kg (c) 11 1 9 % less 19. (c) x = 20 days, y = Rs. 2500 (d) 40 days Project Work Collect 5 problems in our daily life where quadratic equation is applied. Show their solution and present them in your classroom.
Oasis School Mathematics-10 185 10.2 Simplification The process of reducing an expression to the simplest form is known as simplification. Generally, simplification of rational expressions can be done by taking the L.C.M. of the denominators. The resulting terms should be reduced into the simplest form. Let’s see the following examples. I. Simplification of algebraic II. Simplification of algebraic fractions fractions having a common denominator. having different denominators. x y x y x y x y ² ² – ² – ² – + + x x x x + + + − 3 4 1 3 – = x y x y ² ² x ² – y ² – + + = ( )( – ) – ( )( ) ( )( – ) x x x x x x + + + + 3 3 1 4 4 3 = 2x x y ² – = = = = − + + ( ) ( )( – ) 5 13 4 3 x x x ( ² ) ( ² ) ( )( – ) x x x x x − − + + + 9 5 4 4 3 − − + 5 13 4 3 x ( ) x x( – ) x x x x x 2 2 9 5 4 4 3 − − − − ( ) + ( – ) Unit 10 Algebraic Fraction 10.1 Warm-up Activities • Give some examples of algebraic fraction. • What is the L.C.M. of (x+2) and (x+3)? • Simplify: 1 x–4 + 1 x+4 • Simplify: 1 x+2 + 1 x+3 • For what value of x is, 1 x+2 not defined? • For what value of x is, 2x+1 x+1 not defined? Discuss the following in your class and draw a conclusion.
186 Oasis School Mathematics-10 Worked Out examples example 1 Simplify: a b a b a b a b ab a b – – + – ² ² + + + 4 Solution: a b a b a b a b ab a b – – + – ² ² + + + 4 = ( ² – ²) – ( ² ²) ² ² ² ² a ab b a ab b a b ab a b 2 2 + + + 4 − + + = – ² ² ² ² 4 4 ab a b ab − a b + + = – ( ² ²) ( ² – ²) ( ² ²)(a² b²) 4 4 ab a b ab a b a b + + − + = – a b – ab – 4 4 4 4 3 3 3 3 4 4 a b ab a b − + = - - 8 3 4 4 ab a b example 2 Simplify: 2 4 2 2 4 2 16 16 4 3 4 4 x y x xy y x y x xy y x x x y y − − + − + + + + ² ² ² ² + + ² ² Solution: = 2 4 2 2 4 2 16 16 4 3 4 4 x y x xy y x y x xy y x x x y y − − + − + + + + ² ² ² ² + + ² ² = = [( )³ ³] [( x)³ y³] ( ² ²)( x ² xy y ²) ³ ( ² ²) 2 2 4 2 4 2 16 4 x y x xy y x x y − − + − + + + + + ² ( − 2 y x )² = 8 8 4 2 4 2 16 4 2 4 x ³ ³ x³ y³ ( ² ²)( x ² xy y ²) ³ ( ² ²)( x ² − − − − + + + + + + y x xy y x x xy y + − y ² 2 y x ) = − − + + + + + + − + 2 4 2 4 2 16 4 2 4 2 y x xy y x x xy y y y ³ ( ² ²)( x ² xy y ²) ³ ( ² ²) ( x ² x ²) = 16 2 4 2 4 2 x x xy y ³ y³ ( ² ²)( x ² xy y ²) − − + + + = 2 8 4 2 4 2 ( ³ y ³) ( ² ²)( x ² xy y ²) x x xy y − − + + + = 2 2 4 2 4 2 4 2 ( – )( ² ²) ( ² ²)( x ² xy y ²) x y x xy y x xy y + + − + + + = 2 2 4 2 ( – ) ( ² ²) x y x x − + y y ( )( ² ²) – ( )( ² – ²) ( ² ²)( x ² xy 2 4 2 2 4 2 4 2 4 2 x y x xy y x y x xy y x xy y − + + + + − + + + y ²) ³ ( ²)² . ². ² ( ²)² – x ² y ² + + + 16 4 2 4 4 x x x y y
Oasis School Mathematics-10 187 example 3 Simplify: ( – )² ² ² – (y z)² ( )² ² ² ( )² (z )² y ² ² – ( )² x y z x y z x y z x x z x y − + + − − − + + − − + Solution: ( – )² ² ² – (y z)² ( )² ² ² ( )² (z )² y ² ² – ( )² x y z x y z x y z x x z x y − + + − − − + + − − + = ( – )( ) ( y z)(x y z) ( )( ) ( )(y z ) x y z x y z x y z x y z x y z x x + − − + + − − + − + − − + + − − + (z y)(z x y) ( )(z ) − + − − + + − − x z x y x y = x y z x y z x x y z x x y z – y z + z y + + + − + + + + − + + + = x y z y z x z x y x – ( y z) + + − + + − + + + = ( ) ( y z) x y z x + + + + = 1 example 4 Simplify: 1 2 4 8 4 4 7 8 8 a x a a x a a x a + a x + + + + − ² ² − 4a7³ + 4a3 x4 a8 – x8 Solution: = 1 2 4 8 4 4 7 8 8 a x a a x a a x a + a x + + + + − ² ² − 4a7³ + 4a3 x4 a8 – x8 = 1 2 4 3 4 4 4 4 4 4 a x a a x a a x + a x a x + + − + ² ² + − ( ) ( )( ) = 1 2 4 3 4 4 a x a a x a + a x + + − ² ² − = 1 2 4 3 a x a a x a + a x + + ( ² – ²) – ( ² ²)(a² – x ²) = 1 2 2 4 3 a x a ax + a x + − + ³ – ² a ( ² ²)(a² – x ²) = 1 2 a x a + a x + + + – (a ² x ²) ( ² ²)(a² – x ²) = 1 2 a x a + a x − ( + )(a – x) = a x a a x − − + − 2 ( )(a x) = − + + − ( ) ( )(a x) a x a x = - - 1 (a x) 1 x–a =
188 Oasis School Mathematics-10 example 5 Solution: = 1 a2 (a2 –x2 ) 1 a2 (x2 +a2 ) 2a4 (x8 –a8 ) + + 1 a2 (x2 +a2 ) 1 a2 (x2 –a2 ) 2a4 (x8 –a8 ) + + = 1 a2 a2 +x2 +a2 –x2 (a2 –x2 ) (a2 +x2 ) 2a4 (x8 –a8 ) [ ] + = 1 a2 1 a2 –x2 1 a2 +x2 2a4 (x8 –a8 ) [ + ] + = = = = 1 a2 2a2 a4 – x4 2 a4 – x4 2a4 (x8 –a8 ) 2a4 (x4 –a4 ) (x4 +a4 ) 2(a4 +x4 )–2a4 (a4 –x4 ) (a4 +x4 ) 2a4 + 2x4 –2a4 (a4 –x4 ) (a4 +x4 ) [ ] + + = 2x4 a8 – x8 example 6 (a) Factorise the algebraic expression: 16a4 + 4a²b² + b4 . (b) Find the L.C.M of : 4a² + 2ab + b², 4a² – 2ab + b² and 16a4 + 4a²b² + b4 . (c) Simplify the algebraic fraction. 2a + b 4a² + 2ab + b² + 2a – b 4a² – 2ab + b² – 2b³ 16a4 + 4a²b² + b4 Solution: (a) 16a4 + 4a²b² + b4 = (4a²)² + (b²)² + 4a²b² = (4a² + b²)² – 2.4a².b² + 4a²b² = (4a² + b²)² – 4a²b² = (4a² + b²)² – (2ab)² = (4a² + b² + 2ab) (4a² + b² – 2ab) (b) L.C.M. of 4a² + 2ab + b², 4a² – 2ab + b² and 16a4 – 4a²b² + b4 is (4a² + 2ab + b²) (4a² – 2ab + b²) (c) 2a + b 4a² + 2ab + b² + 2a – b 4a² – 2ab + b² – 2b³ 16a4 + 4a²b² + b4 = (2a + b) (4a² – 2ab + b²) + (2a – b) (4a² + 2ab + b²) (4a² + 2ab + b²) (4a² – 2ab + b²) – 2b³ (4a² + 2ab + b²) (4a² – 2ab + b²) [ From (a)]
Oasis School Mathematics-10 189 = (2a)³ + b³ + (2a)³ – b³ (4a² + 2ab + b²) (4a² – 2ab + b²) – 2b³ (4a² + 2ab + b²) (4a² – 2ab + b²) = 16a³ (4a² + 2ab + b²) (4a² – 2ab + b²) – 2b³ (4a² + 2ab + b²) (4a² – 2ab + b²) = 16a³ – 2b³ (4a² + 2ab + b²) (4a² – 2ab + b²) = 2[8a³ – b³] (4a² + 2ab + b²) (4a² – 2ab + b²) = 2 [(2a)³ – b³] (4a² + 2ab + b²) (4a² – 2ab + b²) = 2 (2a – b) (4a² + 2ab + b²) (4a² + 2ab + b²) (4a² – 2ab + b²) = 2 (2a – b) (4a² – 2ab + b²) exercise 10.1 1. (a) On what condition is the rational fraction p(x) q(x) not defined? (b) For what value of x, is 1 2x–1 not defined? (c) What value of x makes the expression 5x+1 2x–6 undefined? (d) On what condition is p(x) q(x)–r(x) not defined? 2. Simplify: 3. Simplify the following: (a) a b a b a b a b ab a b + − − − + + − 2 2 2 (b) 1 1 1 1 4 1 − + − + − + − x x x x x x² (c) x y x y x y x y x y x + − − − + + + − 2( ² ²) ² y ² (d) a b a b a b a b ab b + − − − + − + 4 a ² ² ÷ ÷ + + - ÷ (a) (b) (d) (e) (c) x2 –xy x2 +xy x2 –xy x2 +xy x2 y(x–y) x2 +y2 xy x2 –xy+y2 x–y x2 (x–y) x3 +x2 y x2 (x–y) x3 +x2 y y2 x(y-x) x2 y(x+y) y2 x(x+y) x2 +xy+y2 x+y
190 Oasis School Mathematics-10 Note: 4. Simplify: (a) 2 2 3 2 1 3 1 (x– )(x− ) (x )( ) ( ) 1 2( ) + − − + x x − − x (b) 2 3 4 5 1 3 4 2 5 3 ( ) (a– )(a ) ( ) ( )(a ) ( )(a ) a a a a a − − + − − − + − − − (c) 1 3 2 3 2 4 2 (a a − + )(a ) (a )( ) ( ) a a 3 4 ( ) + + − + − − (d) 1 1 1 (x y − − )(z x z ) (y x)(y ) (z x y )(z ) + − − + − − 5. Simplify: (a) 1 5 6 1 3 2 2 x ² − +x x ² x ² 4 3 + − + − x x − + (b) (c) (d) 6. Simplify: (a) a b ab b a b b a a b b − − + + + + + + a ² ² a ² ab b² + + ³ ² ² 2 4 4 (b) (c) (d) (e) 7. Simplify: (a) ( ) ² ( )² (a )² ² ² (a )² ( )² – ² ² ( )² ( )² a a b c c b b c a b c c a b b c a − − + − + − − + + − − + − ² ( ) ( ) ( a) ( ) b a a a 1 4 1 1 4 1 1 2 1 1 1 2 4 + + − + + − (b) − 4 3 4 4 2 9 9 4 2 2 3 16 a b c 16 2 c a b b c a a b c ² ( )² c ( )² ² ² ( )² ( )² – ² − − ² ( + − + − − + + − a b b c a − + − 3 3 4 4 )² ( )² ² 8. Simplify: x x x x x x x − − + + − − + + − − + 1 3 2 2 5 6 5 ² x ² x ² 8 15 1 10 21 1 12 35 1 x x ² x − + ² x x x ² 8 15 + − + + − + x2 +2xy–8y2 x2 –4y2 x2 –4y2 x2 –2xy x x+2y – + 2 4 2 2 4 2 2 16 4 4 4 a b ab b a b b a a b b + + + + − − + − a ² ² a ² ab b² + + ³ ² ² 3 1 9 3 1 3 1 9 3 1 2 81 9 1 4 a a a a a + + + + − − + − a ² a ² a ² + + x x x x x x + + + + − − + − + + 3 3 9 3 3 9 54 9 81 4 x ² x ² ² a a a a a a a a a + + + − − − + − + + 2 1 2 1 2 1 4 ² ² ² ² 2 (a+1) (a+2) 1 2(1–a2 ) 1 (a+2)2 1 (a+1)2 (a+2)2 1 2(1+a2 ) 1 1-a4 (a) – (b) + + – a 2(a2 –b2 ) a 2(a2 +b2 ) a3 a4 –b4 (c) + – +
Oasis School Mathematics-10 191 9. Simplify: a ac a c c a c c c ² a ²c – c ³ ( ) a ² ² + − − + − − 2 10. Simplify: (a) (b) (c) (d) (e) 11. Simplify : (a) (b) 12. Simplify : (a) (b) 13. Simplify : (a) (b) 14. (a) Reduce the algebraic expression 4a7 + 4a³b4 a8 – b8 into its simplest form. (b) Using the result obtained in (a), simplify 2a a2 – b2 – 4a7 + 4a³b4 a8 – b8 . (c) Using the result obtained in(b), simplify. 1 a + b + 2a a2 – b2 – 4a7 + 4a³b4 a8 – b8 15. (a) Factories the algebraic expression. x² – 3x + 2, x² – x – x and x² – 1. (b) What is their L.C.M? (c) Simplify: 2 x² – 3x + 2 + 2a x² – x – 2 – 4 x² – 1 16. (a) Using the formula of a² – b², reduce the expression (a – b)² + 4ab – c² (a + b – c) into its simplest form. (b) Following the same method, factories the expression (b + c)² – 4bc – a² b – c + a and 4b² + bc 2b + c into their simplest form. (c) Simplify the expression. (a – b)² + 4ab – c² a + b – c + (b + c)² – 4bc – a² b – c + a + 4b² + bc 2b + c 4ax x2 –a2 x a x a x a x a ax x a x x a + − − − + − + − + 4 84 4 ² a ² 1 ³ 1 1 1 4 1 8 1 4 − + − + − − + − + a a a a a a a ² a 4a a2 + 1 2x x2 –4y2 1 2 1 2 2 4 4 16 4 4 x y x y x x y x − x y + + + + + ² ² + ³ 2ab a2 – b 1 2 + + 2 44 4 + − + + − + + + b a b ab a b a a ² ² a b a b ³ b 1 2 44 4 + − + + − + + + b a b ab a b a a ² ² a b a b ³ b 1 1 1 1 1 1 + + − − + − − y x x y y x y x x y y x ² ² ² ² ² ² 1 1– y2 x2 1 1 1 1 4 1 8 1 3 4 + − + − + + + + − x x x x x x x ² x 4x + 4x3 1 – x4 x y x y x y x y xy x x y x y + − + − + + + − − 4 8 3 4 4 ² y ² 4x3 y+4xy3 x4 –y4 4x4 + 4 x8 –1 2 1 1 1 3 + 1 1 + − + − + a + a a a a ² a³ 3a – 1 a2 – 1 1 1 2 1 4 1 8 1 4 8 x x + x x + + + + + ² − 4a7 + 4a3 b4 a8 – b8 2 1 3 3 x y x y y y x xy + x y − − − − + ² ² + ³ x – 3y x2 – y2 1 2 4 8 4 4 7 8 8 a b a b a a b a + a b + + + + − a ² ² − ³
192 Oasis School Mathematics-10 Project Work Collect 5 different types of questions of algebraic fractions. Use the tool of simplification for each type and present it in your classroom. Answers 1. Consult your teacher 2. Consult your teacher 3. (a) (b) 0 6 2 2 ab a b− (c) 2(x+y) (x-y) (d) 8ab a -b 4. (a) 1 (x-2)(x-3) (b) 5 (a-3)(a-4)( 3 4 4 a-5) (c) 9 (a+2)(a-3)(a-4) (d) 2 (x-y)(z-x) 5. ) 0 (b) 3x-7 (x-2)(x (a -3) (c) 3 (x-3)(x-7) 6. (a) 2(a+b) a +ab+b (b) 2(2a-b) 4a 2 2 2 ( ) d x y x +2 -2ab+b (c) 2(3a-1) 9a -3a+1 (d) 2(x-3) x -3x+9 (e) 4 1+a +a 7. (a) 2 2 2 2 4 16a 1-a (b) 16a x x -a (c) 8x x -256y (d) 8a b a -b (e) x y x - 8 7 8 8 7 8 8 7 8 8 2 2 4 5 y (f) 4x 1-x (g) 2m m -1 (h) 0 (i) 4a a +b (j) 4a 4+a 8. (a) 2(1+x) 4 2 4 4 2 2 2 2 2 (1-x) (b) 2(x-y) (x+y) 9. (a) 1 x-1 (b) 1 b-a 10. (a) 1 (1-a)(1+a ) (b) x 3 3 3 3 2 2 2 2 2 (x-y)(x +y ) 11. (a) x +1 (b) 2x (c) 32x (x -9)(x -16) (d) 6a x (x2 2 2 2 -a )(4x -a ) 12. (a)1 (b) 1 13. 14. 3 a+c 15. 3 x ••• •• •• • • • ••• 7. (a) 1 (b) 1 8. (a) , (b) 0, (c) ab2 a4 –b4 1 (a+1)2 9. 3 a+c 10. 2(1+x) (1–x) 1 x–1 1 b–a 2(x-y) (x+y) 11. (a) (b) 12. (a) (b) 1 (1-a) (1+a3 ) x3 (x-y) (x3 +y3 ) 13. (a) (b) 1. Consult your teacher 2. Consult your teacher 3. (a) (b) 0 6 2 2 ab a b− (c) 2(x+y) (x-y) (d) 8ab a -b 4. (a) 1 (x-2)(x-3) (b) 5 (a-3)(a-4)( 3 4 4 a-5) (c) 9 (a+2)(a-3)(a-4) (d) 2 (x-y)(z-x) 5. ) 0 (b) 3x-7 (x-2)(x (a -3) (c) 3 (x-3)(x-7) 6. (a) 2(a+b) a +ab+b (b) 2(2a-b) 4a 2 2 2 ( ) d x y x +2 -2ab+b (c) 2(3a-1) 9a -3a+1 (d) 2(x-3) x -3x+9 (e) 4 1+a +a 7. (a) 2 2 2 2 4 16a 1-a (b) 16a x x -a (c) 8x x -256y (d) 8a b a -b (e) x y x - 8 7 8 8 7 8 8 7 8 8 2 2 4 5 y (f) 4x 1-x (g) 2m m -1 (h) 0 (i) 4a a +b (j) 4a 4+a 8. (a) 2(1+x) 4 2 4 4 2 2 2 2 2 (1-x) (b) 2(x-y) (x+y) 9. (a) 1 x-1 (b) 1 b-a 10. (a) 1 (1-a)(1+a ) (b) x 3 3 3 3 2 2 2 2 2 (x-y)(x +y ) 11. (a) x +1 (b) 2x (c) 32x (x -9)(x -16) (d) 6a x (x2 2 2 2 -a )(4x -a ) 12. (a)1 (b) 1 13. 14. 3 a+c 15. 3 x ••• •• •• • • • ••• 14. (c) 1 15. (b) (x +1)(x – 1) (x – 2) (c) 8 (x² – 1) (x – 2) 16. (c) 4b
Oasis School Mathematics-10 193 11.1 exponential equations Let's consider the terms 5x , 10x , ax , etc. In each of the above terms, the base is constant and power is variable. Such terms are experiential terms. Equations involving exponential terms are called exponential equation like. 2x+5 = 42x + 1 3x+2 + 3x = 30, etc. • ax = a2 ⇒ x = 2 i.e. to equate the indices on both sides of the exponential equation, the bases should be maintained the same, here ‘a’ is any number other than zero. • 24 = x4 ⇒ x = 2 i.e. to equate the bases on both sides of the exponential equation, the powers should be maintained the same. • This condition may not always hold, for example (2)4 = (–2)4 but 2 ≠ –2 Worked Out examples example 1 Solve : 27x = 9(x + 4) Solution: Here, 27x = 9(x + 4) or, (33 )x = 32(x + 4) or, 33x = 32x + 8 3x = 2x + 8 or, 3x – 2x = 8 ∴ x = 8 example 2 Solve: 2x – 2 + 23 + x = 33 Solution: Here, 2x – 2 + 23 + x = 33 or, 2x .2–2 + 23 .2x = 33 or, 2 1 4 8 2 1 32 4 x x + ( ) + ( ) = 33 or, 2 1 4 8 2 1 32 4 x x + ( ) + ( ) = 33 or, 2x (33 4 ) = 33 or, 2x = 22 ∴ x = 2 Substitute x = 8 in the given equation and check whether it satisfies the given equation or not. Unit 11 Indices
194 Oasis School Mathematics-10 example 3 Solve: 3x + 1 3x = 3 1 3 Solution: Here, 3x + 1 3x = 3 1 3 Let, 3x = a ∴ Above equation reduces into a + 1 a = 10 3 or, a2 +1 a = 10 3 or, 3a2 + 3 = 10a or, 3a2 – 10a + 3 = 0 or, 3a2 – 9a – a + 3 = 0 or, 3a (a – 3) – 1 (a – 3) = 0 or, (a – 3) (3a – 1) = 0 Either, a – 3 = 0 or, 3a – 1 = 0 or, a = 3 3a = 1 or, 3x = 31 or, a = 1 3 ∴ x = 1 or, 3x = 3–1 ∴ x = –1 Hence, x = ± 1 example 4 Solve: 9x – 10 × 3x + 9 = 0 Solution: Here, 9x – 10×3x + 9 = 0 or, (32 )x – 10.3x + 9 = 0 or, (3x )2 – 10.3x + 9 = 0 Let, 3x = a, then ∴ a2 – 10a + 9 = 0 or, a2 – 9a – a + 9 = 0 or, a(a – 9) – 1 (a – 9) = 0 or, (a – 9) (a – 1) = 0 Either a – 9 = 0 or, a – 1 = 0 or, a = 9 or, a = 1 or, 3x = 32 or, 3x = 30 ∴ x = 2 ∴ x = 0 Hence, x = 2, 0 example 5 If a = 10x , b = 10y and ay bx = 100, prove that xy = 1. Solution: Here, Given, a = 10x , b = 10y and ay bx = 100 Check whether both values satisfy the given equation or not.
Oasis School Mathematics-10 195 Here, ay .bx = 100 or, (10x )y . (10y )x = 102 or, 10xy. 10xy = 102 (10)2xy = 102 ∴ 2xy = 2 or, xy = 1 proved. example 6 (a) If 3x = 1, what type of equation is this? What is its solution? (b) Solve the equation 4 × 3x + 1 – 27 = 9x . (c) Identify whether solution obtained in (b) satisfies the given equation or not? Solution: (a) 3x = 1 is an exponential equation. Now, 3x = 30 x = 0. (b) Given equation 4 × 3x + 1 – 27 = 9x or, 4 × 3x × 31 – 27 = (32 )x or, 12 × 3x – 27 = (3x )² or, (3x )² – 12 × 3x + 27 = 0 Let, 3x = a or, a² – 12a + 27 = 0 or, a² – (9 + 3)a + 27 = 0 or, a² – 9a – 3a + 27 = 0 or, a(a –9) – 3(a – 9) = 0 or, (a – 9) (a – 3) = 0 Either, a – 9 = 0 or, a – 3 = 0 a = 9 or, a = 3 3x = 3² or, 3x = 31 ∴ x = 2 or x = 1. (c) Checking When x = 2, 4 × 3x + 1 – 27 = 9x or, 4 × 32 + 1 – 27 = 9² or, 4 × 3³ – 27 = 81 or, 108 – 27 = 81 or, 81 = 81 Hence, x = 2 satisfies then given equation.
196 Oasis School Mathematics-10 Again, when x = 1 4 × 31+1 – 27 = 91 4 × 3² – 27 = 9 36 – 27 = 9 9 = 9 (true) Hence, x = 1, also satisfies the equation. exercise 11.1 1. (a) If x2a+1 = (x3 )a , find the value of ‘a’. (b) If ax = b2x, what is the relation of ‘a’ and ‘b’? (c) If 32x–1 = 1, find the value of x. (d) If 5x+1 = 1, find the value of x. (e) If 25x+1 = 26x–1, find the value of x. 2. Solve: (a) 2x = 64 (b) 4x = 1 64 (c) 2x–3 = 16 (d) x-2 = 1 16 (e) 8-x/ 3 = 1 2 (f) 27 x/ 3 = 3 3. Solve the following equations: 4. Solve the following: (a) 23x – 5. ax – 2 = 2x – 2.a1 – x (b) 75x–4. a4x–3 = 72x – 3.ax –2 (c) 2x – 2 + 2x = 5 (d) 3x + 2 + 3x+3 3 = 2 (e) 5x + 5x + 1 + 5x + 2 = 155 (f) 3x+3 + 3x = 28 (g) 3x–2+ 3x = 10 9 (h) 32x + 3 – 2×9x+1 = 1 3 (i) 2x–5 × 5x–4 = 5 (j) 2x+3 × 3x+4 = 18. 5. Solve: (a) 4x – 3×2x + 2 = 0 (c) 2x + 1 2x = 4 1 4 ( ) ( ) ( ) ( ) a b c d x x x x x x x 3 9 2 2 2 2 1 2 16 2 1 2 1 5 3 2 5 1 3 2 3 2 + − + + + + = ( ) =( ) = × ( ) 1 5 1 2 3 3 2 64 5 25 125 10 1 0 001 0 4 0 =( ) × = = = + − x x x y x e f g ( ) ( ) ( ) ( ) . ( ) ( . ) / . ( ) ( ) . ( ) ( ) ( ) ( ) 16 25 1 0 04 2 3 27 8 2 16 1 5 3 2 2 3 2 h i j k x x x x x + − + = ( ) =( ) = =( ) 25 −2