CAPITULO XXXVI
ECUACIONES BINOMIAS Y TRINOMIAS
ECUACIONES BINOMIAS Y TRINOMIAS
- Ejercicio 283
Resolver las ecuaciones siguientes, hallando todas las raΓces:
- x 4 β10 x 2 +9 =0 ( x 2 β9 ) ( x 2 β1 ) =0 ( xβ3 ) ( x+3 ) ( xβ1 ) ( x+1 ) =0 x 1 β3 =0 x 1 =3 x 2 +3 =0 x 2 =β3 x 3 β1 =0 x 3 =1 x 4 +1 =0 x 4 =β1
- x 4 β13 x 2 +36 =0 ( x 2 β4 ) ( x 2 β9 ) =0 ( xβ2 ) ( x+2 ) ( xβ3 ) ( x+3 ) =0 x 1 β2 =0 x 1 =2 x 2 +2 =0 x 2 =β2 x 3 β3 =0 x 3 =3 x 4 +3 =0 x 4 =β3
- x 4 β29 x 2 +100 =0 ( x 2 β25 ) ( x 2 β4 ) =0 ( xβ5 ) ( x+5 ) ( xβ2 ) ( x+2 ) =0 x 1 β5 =0 x 1 =5 x 2 +5 =0 x 2 =β5 x 3 β2 =0 x 3 =2 x 4 +2 =0 x 4 =β2
- x 4 β61 x 2 +900 =0 ( x 2 β36 ) ( x 2 β25 ) =0 ( xβ6 ) ( x+6 ) ( xβ5 ) ( x+5 ) =0 x 1 β6 =0 x 1 =6 x 2 +6 =0 x 2 =β6 x 3 β5 =0 x 3 =5 x 4 +5 =0 x 4 =β5
- x 4 +3 x 2 β4 =0 ( x 2 +4 ) ( x 2 β1 ) =0 ( x 2 +4 ) ( xβ1 ) ( x+1 ) =0 x 1 β1 =0 x 1 =1 x 2 +1 =0 x 2 =β1 x 2 +4 =0 x 2 =β4 x = Β± ( β1 ) ( 2 ) 2 x = Β± 2 β1 x = Β± 2i x 3 =2i x 4 =β2i
- x 4 +16 x 2 β225 =0 ( x 2 +25 ) ( x 2 β9 ) =0 ( x 2 +25 ) ( xβ3 ) ( x+3 ) =0 x 1 β3 =0 x 1 =3 x 2 +3 =0 x 2 =β3 x 2 +25 =0 x 2 =β25 x = Β± ( β1 ) ( 5 2 ) x = Β± 5 β1 x = Β± 5i x 3 =5i x 4 =β5i
- x 4 β45 x 2 β196 =0 ( x 2 β49 ) ( x 2 +4 ) =0 ( xβ7 ) ( x+7 ) ( x 2 +4 ) =0 x 1 β7 =0 x 1 =7 x 2 +7 =0 x 2 =β7 x 2 +4 =0 x 2 =β4 x = Β± β 2 2 x = Β± 2 β1 x = Β± 2i x 3 =2i x 4 =β2i
- x 4 β6 x 2 +5 =0 ( x 2 β5 ) ( x 2 β1 ) =0 ( x 2 β5 ) ( xβ1 ) ( x+1 ) =0 x 1 β1 =0 x 1 =1 x 2 +1 =0 x 2 =β1 x 2 β5 =0 x = Β± 5 x 3 = 5 x 4 =β 5
- 4 x 4 β37 x 2 +9 =0 4 x 4 β x 2 β36 x 2 +9 =0 x 2 ( 4 x 2 β1 ) β9( 4 x 2 β1 ) =0 ( x 2 β9 ) ( 4 x 2 β1 ) =0 ( xβ3 ) ( x+3 ) ( 4 x 2 β1 ) =0 x 1 β3 =0 x 1 =3 x 2 +3 =0 x 2 =β3 4 x 2 β1 =0 4 x 2 =1 x 2 = 1 4 x = Β± 1 4 x = Β± 1 2 1 x 3 = 1 2 1 x 4 =β 1 2 1
- 9 x 4 β40 x 2 +16 =0 9 x 4 β36 x 2 β4 x 2 +16 =0 9 x 2 ( x 2 β4 ) β4( x 2 β4 ) =0 ( 9 x 2 β4 ) ( x 2 β4 ) =0 ( 9 x 2 β4 ) ( xβ2 ) ( x+2 ) =0 x 1 β2 =0 x 1 =2 x 2 +2 =0 x 2 =β2 9 x 2 β4 =0 9 x 2 =4 x 2 = 4 9 x = Β± 4 9 x = Β± 2 3 x 3 = 2 3 x 4 =β 2 3
- 25 x 4 +9 x 2 β16 =0 25 x 4 +25 x 2 β16 x 2 β16 =0 25 x 2 ( x 2 +1 ) β16( x 2 +1 ) =0 ( 25 x 2 β16 ) ( x 2 +1 ) =0 x 2 +1 =0 x 2 =β1 x = Β± β1 x 1 =i x 2 =βi 25 x 2 β16 =0 25 x 2 =16 x 2 = 16 25 x = Β± 16 25 x = Β± 4 5 x 3 = 4 5 x 4 =β 4 5
- 4 x 4 +11 x 2 β3 =0 4 x 4 β x 2 +12 x 2 β3 =0 x 2 ( 4 x 2 β1 ) +3( 4 x 2 β1 ) =0 ( x 2 +3 ) ( 4 x 2 β1 ) =0 4 x 2 β1 =0 4 x 2 =1 x 2 = 1 4 x = Β± 1 4 x = Β± 1 2 x 1 = 1 2 x 2 =β 1 2 x 2 +3 =0 x 2 =β3 x = Β± β3 x = Β± β1 3 x = Β± 3 i x 3 = 3 i x 4 =β 3 i
- ( 2 x 2 +1 ) 2 β ( x 2 β3 ) 2 =80 [ ( 2 x 2 +1 ) β( x 2 β3 ) ] [ ( 2 x 2 +1 ) +( x 2 β3 ) ] =80 [ 2 x 2 +1β x 2 +3 ] [ 2 x 2 +1+ x 2 β3 ] =80 ( x 2 +4 ) ( 3 x 2 β2 ) =80 3 x 4 β2 x 2 +12 x 2 β8 =80 3 x 4 +10 x 2 β88 =0 3 x 4 β12 x 2 +22 x 2 β88 =0 3 x 2 ( x 2 β4 ) +22( x 2 β4 ) =0 ( 3 x 2 +22 ) ( x 2 β4 ) =0 x 2 β4 =0 x = Β± 4 x = Β± 2 x 1 =2 x 2 =β2 3 x 2 +22 =0 3 x 2 =β22 x 2 =β 22 3 x = Β± β 22 3 x = Β± ( β1 ) ( 22 3 ) x = Β± 22 3 i x 3 = 22 3 i x 4 =β 22 3 i
- x 2 ( 3 x 2 +2 ) =4( x 2 β3 ) +13 3 x 4 +2 x 2 =4 x 2 β12+13 3 x 4 +2 x 2 β4 x 2 β1 =0 3 x 4 β2 x 2 β1 =0 3 x 4 β3 x 2 + x 2 β1 =0 3 x 2 ( x 2 β1 ) +( x 2 β1 ) =0 ( 3 x 2 +1 ) ( x 2 β1 ) =0 x 2 β1 =0 x 2 =1 x = Β± 1 x = Β± 1 x 1 =1 x 2 =β1 3 x 2 +1 =0 3 x 2 =β1 x 2 =β 1 3 x = Β± β 1 3 x = Β± ( β1 ) ( 1 3 ) x = Β± 1 3 i x 3 = 1 3 i x 4 =β 1 3 i