CAPITULO XXXVI
ECUACIONES BINOMIAS Y TRINOMIAS
ECUACIONES BINOMIAS Y TRINOMIAS
- Ejercicio 282
Resolver las ecuaciones:
- x 4 β1 =0 ( x 2 +1 ) ( x 2 β1 ) =0 ( x 2 +1 ) ( x+1 ) ( xβ1 ) =0 x 1 +1 =0 x 1 =β1 x 2 β1 =0 x 2 =1 x 2 +1 =0 x 2 =β1 x = Β± β1 x 3 =i x 4 =βi
- x 3 +1 =0 ( x+1 ) ( x 2 βx+1 ) =0 x+1 =0 x 1 =β1 x 2 βx+1 =0 x = β( β1 ) Β± ( β1 ) 2 β4( 1 ) ( 1 ) 2( 1 ) x = 1 Β± 1β4 2 x = 1 Β± β1 3 2 x = 1 Β± 3 i 2 x 1 = 1+ 3 i 2 x 2 = 1β 3 i 2
- x 4 =81 x 4 β81 =0 ( x 2 +9 ) ( x 2 β9 ) =0 ( x 2 +9 ) ( xβ3 ) ( x+3 ) =0 x 1 β3 =0 x 1 =3 x 2 +3 =0 x 2 =β3 x 2 +9 =0 x 2 =β9 x = Β± β 3 2 x = Β± 3 β1 x 3 =3i x 4 =β3i
- x 4 β256 =0 ( x 2 +16 ) ( x 2 β16 ) =0 ( x 2 +16 ) ( xβ4 ) ( x+4 ) =0 x 1 β4 =0 x 1 =4 x 2 +4 =0 x 2 =β4 x 2 +16 =0 x 2 =β16 x = Β± β 4 2 x = Β± 4 β1 x 3 =4i x 4 =β4i
- x 3 +8 =0 ( x+2 ) ( x 2 β2x+4 ) =0 x 1 +2 =0 x 1 =β2 x 2 β2x+4 =0 x = β( β2 ) Β± ( β2 ) 2 β4( 1 ) ( 4 ) 2( 1 ) x = 2 Β± 4β16 2 x = 2 Β± β12 2 x = 2 Β± ( β3 ) ( 2 2 ) 2 x = 2 Β± 2 ( 3 ) ( β1 ) 2 x = 2 ( 1 Β± 3 i ) 2 x 2 =1+ 3 i x 3 =1β 3 i
- x 4 β625 =0 ( x 2 +25 ) ( x 2 β25 ) =0 ( x 2 +25 ) ( xβ5 ) ( x+5 ) =0 x 1 β5 =0 x 1 =5 x 2 +5 =0 x 2 =β5 x 2 +25 =0 x 2 =β25 x = Β± β25 x = Β± ( β1 ) ( 5 2 ) x = Β± 5i x 3 =5i x 4 =β5i
- x 3 +64 =0 ( x+4 ) ( x 2 β4x+16 ) =0 x+4 =0 x 1 +4 =0 x 1 =β4 x 2 β4x+16 =0 x = β( β4 ) Β± ( β4 ) 2 β4( 1 ) ( 16 ) 2( 1 ) x = 4 Β± 16β64 2 x = 4 Β± β48 2 x = 4 Β± ( β3 ) ( 4 2 ) 2 x = 4 Β± 4 3 β1 2 x = 2 ( 2 Β± 2 3 i ) 2 x 2 =2+2 3 i x 3 =2β2 3 i
- x 6 β729 =0 ( x 3 β27 ) ( x 3 +27 ) =0 ( xβ3 ) ( x 2 +3x+9 ) ( x+3 ) ( x 2 β3x+9 ) =0 x 1 β3 =0 x 1 =3 x 2 +3 =0 x 2 =β3 x 2 +3x+9 =0 x = β3 Β± 3 2 β4( 1 ) ( 9 ) 2( 1 ) x = β3 Β± 9β36 2 x = β3 Β± β27 2 x = β3 Β± ( β1 ) ( 3 3 ) 2 x = β3 Β± 3 3 β1 2 x = β3 Β± 3 3 i 2 x 3 = β3β3 3 i 2 x 4 = β3+3 3 i 2 x 2 β3x+9 =0 x = β( β3 ) Β± ( β3 ) 2 β4( 1 ) ( 9 ) 2( 1 ) x = 3 Β± 9β36 2 x = 3 Β± β27 2 x = 3 Β± ( β1 ) ( 3 3 ) 2 x = 3 Β± 3 3 β1 2 x = 3 Β± 3 3 i 2 x 5 = 3β3 3 i 2 x 6 = 3+3 3 i 2
- Hallar las raΓces cΓΊbicas de 8.
x 3 =8 x 3 β8 =0 ( xβ2 ) ( x 2 +2x+4 ) =0 x 1 β2 =0 x 1 =2 x 2 +2x+4 =0 x = β2 Β± 2 2 β4( 1 ) ( 4 ) 2( 1 ) x = β2 Β± 4β16 2 x = β2 Β± β12 2 x = β2 Β± ( β1 ) ( 3 ) ( 2 2 ) 2 x = β2 Β± 2 β1 3 2 x = 2 ( β1 Β± 3 i ) 2 x 2 =β1+ 3 i x 3 =β1β 3 i - Hallar las raΓces cuartas de 64.
x 4 =64 x 4 β64 =0 ( x 2 +8 ) ( x 2 β8 ) =0 x 2 β8 =0 x 2 =8 x = Β± 8 x = Β± 2 3 x = Β± 2 2 x 1 =2 2 x 2 =β2 2 x 2 +8 =0 x 2 =β8 x = Β± β8 x = Β± β 2 3 x = Β± 2 2 β1 x 1 =2 2 i x 2 =β2 2 i