Tiêu đề Polynomial Engg Question Bank Solution.pdf ✅

2  Higher Math 2nd Paper Chapter-4  +  +  = – 14  6 + 4 ( + ) = – 14   +  = – 5 q = – 5  +  = 2   = – 2  = 6  = – 3 r =  = (– 2) × 4 = – 8 p =  +  +  = – 1  p + q + r = – 1 – 5 – 8 = – 14 6| ‘a’ Gi ev ̄Íe gvb KZ n‡j x 3 + 3ax2 + x + 1 = 0 mgxKi‡Yi g~j ̧‡jv mgvšÍi cÖMg‡b _vK‡e? mgxKiYwUi g~j ̧‡jvI wbY©q Ki| [BUET 14-15] mgvavb: awi, g~j ̧‡jv: ( – d), , ( + d)  3 = – 3a   = – a ......... (i) Avevi, ( – d) () ( + d) = – 1  ( 2 – d 2 ) = – 1   2 – d 2 = – 1  ......... (ii) Avevi, ( – d) + .( + d) +  2 – d 2 = 1  2 2 + ( 2 – d 2 ) = 1  2 2 + – 1  = 1  2 3 –  – 1 = 0   = 1 (i) bs n‡Z, a = – 1 (Ans.) (ii) n‡Z, 1 – d 2 = – 1 d = 2 g~j ̧‡jv 1 – 2 , 1, 1 + 2 (Ans.) 7| hw` x 3 – px2 – qx – r = 0 mgxKi‡Yi g~j ̧wj a, b, c nq Z‡e a 3 + b3 + c3 Gi gvb wbY©q Ki| [BUET 14-15] mgvavb: a + b + c = p ab + bc + ca = – q abc = r a 3 + b3 + c3 = (a + b + c) {(a + b + c)2 – 3(ab + bc + ca)} + 3abc = p{(p)2 – 3(– q)} + 3r = p(p2 + 3q) + 3r (Ans.) 8| hw` x 2 + px + q = 0, x2 + qx + 8p = 0 Ges 4x3 + 16x2 – 9x – 36 = 0 mgxKiY ̧‡jvi GKwU mvaviY g~j _v‡K Ges 4x3 + 16x2 – 9x – 36 = 0 mgxKi‡Yi Ab ̈ `yBwU g~‡ji †hvMdj k~b ̈ n‡j, p Ges q Gi gvb wbY©q Ki| [BUET 13-14] mgvavb: awi, 4x3 + 16x2 – 9x – 36 = 0 mgxKi‡Yi g~j ̧‡jv , – ,   –  +  = – 16 4 = – 4  = – 4 x 2 + px + q = 0 Ges x 2 + qx + 8p = 0 Gi mvaviY g~j,  = – 4 16 – 4p + q = 0 ......... (i) 16 – 4q + 8p = 0 ......... (ii) (i) Ges (ii) bs mgxKiY mgvavb K‡i, p = 10; q = 24 (Ans.) 9| 2x2 + 3x + 5 = 0 mgxKi‡Yi g~jØq  Ges  n‡j 1  3Ges 1  3 g~jØq Øviv MwVZ mgxKiYwU wbY©q Ki| [BUET 05-06] mgvavb:  +  = – 3 2 ,  = 5 2 ; wb‡Y©q mgxKiY, x 2 –     1  3 + 1  3 x + 1  3  3 = 0  x 2 –      ( + )  3 – 3( + ) () 3 x + 1  3  3 = 0  x 2 –     – 3 2 3 – 3 × 5 2     – 3 2     5 2 3 x + 1     5 2 3 = 0 wb‡Y©q mgxKiY: x 2 – 63 125 x + 8 125 = 0  125x2 – 63x + 8 = 0 (Ans.) 10| k Gi gvb KZ n‡j (k2 – 3)x2 + 3kx + (3k + 1) = 0 mgxKi‡Yi GKwU g~j Aci g~jwUi D‡ëv n‡e? AZtci mgxKi‡Yi g~j؇qi ag© wbY©q Ki| [BUET 04-05] mgvavb: g~jØq Dëv n‡j, k 2 – 3 = 3k + 1  k 2 – 3k – 4 = 0  k = 4, – 1 k = – 1 n‡j mgxKiYwU n‡e – 2x2 – 3x – 2 = 0 D = 32 – 4.2.2 = – 7 < 0  g~jØq AbyeÜx RwUj| Avevi, k = 4 n‡j mgxKiY n‡e 13x2 + 12x + 13 = 0 D = 122 – 4  13  13 = – 532 < 0  g~jØq AbyeÜx RwUj| 11| 27x2 + 6x – (p + 2) = 0 Gi GKwU g~j AciwUi e‡M©i mgvb n‡j p Gi gvb †ei Ki| [BUET 03-04] mgvavb: cÖ`Ë mgxKiY, 27x2 + 6x – (p + 2) = 0 mgxKi‡Yi GKwU g~j  n‡j Aci g~j  2 ; AZGe,  +  2 = – 6 27 27 2 + 27 + 6 = 0   = – 1 3 , – 2 3 Ges . 2 = – (p + 2) 27   3 = – (p + 2) 27  = – 1 3 n‡j, – 1 27 = – (p + 2) 27  p = – 1 Avevi,  = – 2 3 n‡j, – 8 27 = – (p + 2) 27  p = 6  p = 6 A_ev – 1 (Ans.)

4  Higher Math 2nd Paper Chapter-4     1  + 3       3  + 1  = ( + 3) (3 + ) () 2 = 3 2 +  + 9 + 3 2 () 2 = 3( + ) 2 + 4 () 2 = 3  25 49 – 3 7  4 9 49 = –1  mgxKiY, 3x2 + 20x – 3 = 0 (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 19|  I , x 2 – bx – b = 0 Gi `yBwU g~j|  4 I  4 g~jØq wewkó mgxKiYwU †ei Ki| [RUET 18-19] mgvavb:  +  = b I  = – b   2 +  2 = ( + ) 2 – 2 = b2 + 2b   4 +  4 = ( 2 +  2 ) 2 – 2() 2 = (b2 + 2b)2 – 2b2 = b4 + 4b3 + 4b2 – 2b2  4 +  4 = b4 + 4b3 + 2b2 eqn : x2 – ( 4 +  4 )x + () 4 = 0  x 2 – (b4 + 4b3 + 2b2 )x + b4 = 0 (Ans.) 20| ax2 + bx + c = 0 mgxKi‡Yi g~jØq  I  n‡j (a + b)–2 + (a + b)–2 Gi gvb wbY©q Ki| [RUET 15-16, 07-08] mgvavb: ax2 + bx + c = 0   +  = – b a  a + a = – b Ges  = c a GLb, a + b = – a Ges a + b = – a  (a + b)–2 + (a + b)–2  1 a 2  2 + 1 a 2 2 = ( + ) 2 – 2 a 2 () 2 = b 2 a 2 – 2c a a 2  c 2 a 2 = b 2 – 2ca a 2 c 2 (Ans.) 21| hw` ax2 + bx + c = 0 mgxKi‡Yi g~j `yBwU  Ges  nq, Z‡e ac(x2 + 1) – (b2 – 2ac)x = 0 Gi g~j `yBwU ,  Gi gva ̈‡g cÖKvk Ki| [RUET 12-13] mgvavb: ax2 + bx + c = 0 Gi g~j ,    +  = – b a Ges  = c a ac(x2 + 1) – (b2 – 2ac)x = 0  c a (x2 + 1) –           b a 2 – 2 c a x = 0  x 2 – [{– ( + )}2 – 2]x +  = 0  x 2 – ( 2 +  2 )x +  = 0 x 2 –       +   x +      = 0  x =   ,   (Ans.) 22| ax2 + bx + c = 0 Gi GKwU g~j AciwUi n ̧Y n‡j †`LvI †h, nb2 = ac(1 + n)2 [RUET 10-11] mgvavb: awi, g~jØq  Ges n   + n = – b a  (n + 1)  = – b a   = – b (n + 1)a Avevi, .n = c a  n b 2 (n + 1) 2 a 2 = c a  nb2 = ac(1 + n)2 (Showed) 23| hw` px2 + qx + 1 = 0 Ges qx2 + px + 1 = 0 mgxKiY `yBwUi GKwU gvÎ mvaviY g~j _v‡K, Z‡e cÖgvY Ki †h, p + q + 1 = 0 [RUET 09-10] mgvavb: awi, mvaviY g~j   p 2 + q + 1 = 0 ......... (i)  q 2 + p + 1 = 0 ......... (ii) (i) – (ii)   2 (p – q) + (q – p) = 0   2 (p – q) – (p – q) = 0  (p – q) ( 2 – ) = 0   2 –  = 0 ∵ p  q   2 =    = 1  (i) †_‡K, p + q + 1 = 0 (Proved) 24| hw` px2 + qx + q = 0 mgxKi‡Yi g~j؇qi AbycvZ m : n nq, Z‡e cÖgvY Ki †h, m n + n m + q p = 0 [RUET 08-09] mgvavb: px2 + qx + q = 0 awi, g~jØq m I n  m + n = – q p  (m + n) = –q p Avevi, m.n = q p mn 2 = q p ; mn = q p 2 L.H.S = m n + n m + q p = m + n mn + q p = – q p  p 2 q + q p = – q p + q p = 0 = R.H.S (Proved)