Tiêu đề Polynomial Engg Question Bank Solution.pdf ✅

eûc`x I eûc`x mgxKiY  Engineering Question Bank Solution 3 12| PZzN©vZ wewkó GKwU mgxKiY MVb Ki hvi `ywU g~j h_vμ‡g 2, 3 Ges evKx `ywU g~j x 2 + 4x + 5 = 0 GB mgxKi‡Yi g~j? [BUET 02-03] mgvavb: wb‡Y©q PZzN©vZ mgxKiYwU, (x – 2) (x – 3) (x2 + 4x + 5) = 0 (x2 – 5x + 6) (x2 + 4x + 5) = 0  x 4 – x 3 – 9x2 – x + 30 = 0 (Ans.) 13| hw` x 2 + px + q = 0 Ges x 2 + qx + p = 0 mgxKiY؇qi GKwU mvaviY g~j _v‡K Z‡e 2x2 + (p + q – 2)x = (p + q – 2)2 mgxKi‡Yi g~jØq wbY©q Ki| [BUET 02-03] mgvavb: awi, mvaviY g~j |   2 + p + q = 0 ......... (i)  2 + q + p = 0 .......... (ii) (i) – (ii)  (p – q) + (q – p) = 0   = 1  1 + p + q = 0  p + q = –1 2x2 + (p + q – 2)x = (p + q – 2)2  2x2 – 3x = 9  x = 3, – 3 2 (Ans.) 14| hw`  I  Amgvb nq Ges  2 = 5 – 3 Ges  2 = 5 – 3 nq Z‡e   Ges   g~j wewkó mgxKiYwU wbY©q Ki| [BUET 00-01] mgvavb: cÖkœg‡Z,  I , x 2 – 5x + 3 = 0 mgxKi‡Yi `ywU g~j   +  = 5,  = 3 wb‡Y©q mgxKiY, x 2 –       +   x +   = 0  x 2 –      ( + )  2 – 2  x +   = 0  x 2 – 25 – 6 3 x + 1 = 0 3x2 – 19x + 3 = 0 (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 15| x 4 + 7x3 + 8x2 – 28x – 48 = 0 mgxKi‡Yi mKj g~j wbY©q Ki, hw` `ywU g~‡ji †hvMdj k~b ̈ nq| [KUET 19-20] mgvavb: awi, g~j ̧‡jv a, – a, b, c  b + c = –7 ....... (i) Avevi, abc – abc – a 2 b – a 2 c = 28  – a 2 (b + c) = 28  a 2 = 28 7 = 4  a =  2 Avevi, – a 2 bc = – 48 bc = 12  b – c = (b + c) 2 – 4bc = 49 – 48 = 1 ...... (ii) (i) + (ii)  2b = – 6 b = – 3  c = – 4  g~j ̧‡jv nj 2, – 2, – 3, – 4 (Ans.) 16| hw` x 2 + 2bx + c = 0 mgxKi‡Yi g~jØq  I  nq, n‡e  2 I  2 g~j m¤^wjZ mgxKiYwU wbY©q Ki|  I  Gi gvb I wbY©q Ki| [KUET 06-07] mgvavb:  +  = – 2b  = c  2 +  2 = ( + ) 2 – 2 = 4b2 – 2c   2 I  2 g~j wewkó mgxKiY x 2 – ( 2 +  2 )x +  2  2 = 0  x 2 – (4b2 – 2c)x + c2 = 0  x 2 – 2(2b2 – c)x + c2 = 0  –  = ( + ) 2 – 4 = 4b2 – 4c = 2 b 2 – c   = – b + b 2 – c (Ans.)   = – b – b 2 – c (Ans.) 17| x 2 + bx + c = 0 mgxKi‡Yi g~j `ywU ev ̄Íe I Amgvb n‡j †`LvI †h, 2x2 – 4(1 + c)x + (b2 + 2c2 + 2) = 0 mgxKiYwUi g~j `ywU KvíwbK n‡e| [KUET 05-06] mgvavb: D1 = b2 – 4.c > 0 D2 = 42 (1 + c)2 – 4(b2 + 2c2 + 2)  2 = 16 + 32c + 16c2 – 8b2 – 16c2 – 16 = 32c – 8b2 = – 8(b2 – 4c) < 0 [∵ b 2 – 4c > 0]  2q mgxKiYwUi g~j `yBwU KvíwbK n‡e| (Showed) 18| 7x2 – 5x – 3 = 0 mgxKi‡Yi g~jØq ,  n‡j Giƒc Ges ALÛ mnMwewkó mgxKiY MVb Ki hvi g~j 1  + 3  , 3  + 1  n‡e| [KUET 04-05] mgvavb: GLv‡b,  +  = 5 7  = – 3 7      1  + 3  +     3  + 1  = 4      +   = 4  5 7 – 3 7 = – 20 3