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Ejercicio 283

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CAPITULO XXXVI

ECUACIONES BINOMIAS Y TRINOMIAS
Ejercicio 283
Resolver las ecuaciones siguientes, hallando todas las raΓ­ces:
  1. x 4 –10 x 2 +9 =0 ( x 2 –9 ) ( x 2 –1 ) =0 ( x–3 ) ( x+3 ) ( x–1 ) ( x+1 ) =0 x 1 –3 =0 x 1 =3 x 2 +3 =0 x 2 =–3 x 3 –1 =0 x 3 =1 x 4 +1 =0 x 4 =–1
  2. x 4 –13 x 2 +36 =0 ( x 2 –4 ) ( x 2 –9 ) =0 ( x–2 ) ( x+2 ) ( x–3 ) ( x+3 ) =0 x 1 –2 =0 x 1 =2 x 2 +2 =0 x 2 =–2 x 3 –3 =0 x 3 =3 x 4 +3 =0 x 4 =–3
  3. x 4 –29 x 2 +100 =0 ( x 2 –25 ) ( x 2 –4 ) =0 ( x–5 ) ( x+5 ) ( x–2 ) ( x+2 ) =0 x 1 –5 =0 x 1 =5 x 2 +5 =0 x 2 =–5 x 3 –2 =0 x 3 =2 x 4 +2 =0 x 4 =–2
  4. x 4 –61 x 2 +900 =0 ( x 2 –36 ) ( x 2 –25 ) =0 ( x–6 ) ( x+6 ) ( x–5 ) ( x+5 ) =0 x 1 –6 =0 x 1 =6 x 2 +6 =0 x 2 =–6 x 3 –5 =0 x 3 =5 x 4 +5 =0 x 4 =–5
  5. x 4 +3 x 2 –4 =0 ( x 2 +4 ) ( x 2 –1 ) =0 ( x 2 +4 ) ( x–1 ) ( x+1 ) =0 x 1 –1 =0 x 1 =1 x 2 +1 =0 x 2 =–1 x 2 +4 =0 x 2 =–4 x = Β± ( –1 ) ( 2 ) 2 x = Β± 2 –1 x = Β± 2i x 3 =2i x 4 =–2i
  6. x 4 +16 x 2 –225 =0 ( x 2 +25 ) ( x 2 –9 ) =0 ( x 2 +25 ) ( x–3 ) ( x+3 ) =0 x 1 –3 =0 x 1 =3 x 2 +3 =0 x 2 =–3 x 2 +25 =0 x 2 =–25 x = Β± ( –1 ) ( 5 2 ) x = Β± 5 –1 x = Β± 5i x 3 =5i x 4 =–5i
  7. x 4 –45 x 2 –196 =0 ( x 2 –49 ) ( x 2 +4 ) =0 ( x–7 ) ( x+7 ) ( x 2 +4 ) =0 x 1 –7 =0 x 1 =7 x 2 +7 =0 x 2 =–7 x 2 +4 =0 x 2 =–4 x = Β± – 2 2 x = Β± 2 –1 x = Β± 2i x 3 =2i x 4 =–2i
  8. x 4 –6 x 2 +5 =0 ( x 2 –5 ) ( x 2 –1 ) =0 ( x 2 –5 ) ( x–1 ) ( x+1 ) =0 x 1 –1 =0 x 1 =1 x 2 +1 =0 x 2 =–1 x 2 –5 =0 x = Β± 5 x 3 = 5 x 4 =– 5
  9. 4 x 4 –37 x 2 +9 =0 4 x 4 – x 2 –36 x 2 +9 =0 x 2 ( 4 x 2 –1 ) –9( 4 x 2 –1 ) =0 ( x 2 –9 ) ( 4 x 2 –1 ) =0 ( x–3 ) ( x+3 ) ( 4 x 2 –1 ) =0 x 1 –3 =0 x 1 =3 x 2 +3 =0 x 2 =–3 4 x 2 –1 =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 =– 1 2 1
  10. 9 x 4 –40 x 2 +16 =0 9 x 4 –36 x 2 –4 x 2 +16 =0 9 x 2 ( x 2 –4 ) –4( x 2 –4 ) =0 ( 9 x 2 –4 ) ( x 2 –4 ) =0 ( 9 x 2 –4 ) ( x–2 ) ( x+2 ) =0 x 1 –2 =0 x 1 =2 x 2 +2 =0 x 2 =–2 9 x 2 –4 =0 9 x 2 =4 x 2 = 4 9 x = Β± 4 9 x = Β± 2 3 x 3 = 2 3 x 4 =– 2 3
  11. 25 x 4 +9 x 2 –16 =0 25 x 4 +25 x 2 –16 x 2 –16 =0 25 x 2 ( x 2 +1 ) –16( x 2 +1 ) =0 ( 25 x 2 –16 ) ( x 2 +1 ) =0 x 2 +1 =0 x 2 =–1 x = Β± –1 x 1 =i x 2 =–i 25 x 2 –16 =0 25 x 2 =16 x 2 = 16 25 x = Β± 16 25 x = Β± 4 5 x 3 = 4 5 x 4 =– 4 5
  12. 4 x 4 +11 x 2 –3 =0 4 x 4 – x 2 +12 x 2 –3 =0 x 2 ( 4 x 2 –1 ) +3( 4 x 2 –1 ) =0 ( x 2 +3 ) ( 4 x 2 –1 ) =0 4 x 2 –1 =0 4 x 2 =1 x 2 = 1 4 x = Β± 1 4 x = Β± 1 2 x 1 = 1 2 x 2 =– 1 2 x 2 +3 =0 x 2 =–3 x = Β± –3 x = Β± –1 3 x = Β± 3 i x 3 = 3 i x 4 =– 3 i
  13. ( 2 x 2 +1 ) 2 – ( x 2 –3 ) 2 =80 [ ( 2 x 2 +1 ) –( x 2 –3 ) ] [ ( 2 x 2 +1 ) +( x 2 –3 ) ] =80 [ 2 x 2 +1– x 2 +3 ] [ 2 x 2 +1+ x 2 –3 ] =80 ( x 2 +4 ) ( 3 x 2 –2 ) =80 3 x 4 –2 x 2 +12 x 2 –8 =80 3 x 4 +10 x 2 –88 =0 3 x 4 –12 x 2 +22 x 2 –88 =0 3 x 2 ( x 2 –4 ) +22( x 2 –4 ) =0 ( 3 x 2 +22 ) ( x 2 –4 ) =0 x 2 –4 =0 x = Β± 4 x = Β± 2 x 1 =2 x 2 =–2 3 x 2 +22 =0 3 x 2 =–22 x 2 =– 22 3 x = Β± – 22 3 x = Β± ( –1 ) ( 22 3 ) x = Β± 22 3 i x 3 = 22 3 i x 4 =– 22 3 i
  14. x 2 ( 3 x 2 +2 ) =4( x 2 –3 ) +13 3 x 4 +2 x 2 =4 x 2 –12+13 3 x 4 +2 x 2 –4 x 2 –1 =0 3 x 4 –2 x 2 –1 =0 3 x 4 –3 x 2 + x 2 –1 =0 3 x 2 ( x 2 –1 ) +( x 2 –1 ) =0 ( 3 x 2 +1 ) ( x 2 –1 ) =0 x 2 –1 =0 x 2 =1 x = Β± 1 x = Β± 1 x 1 =1 x 2 =–1 3 x 2 +1 =0 3 x 2 =–1 x 2 =– 1 3 x = Β± – 1 3 x = Β± ( –1 ) ( 1 3 ) x = Β± 1 3 i x 3 = 1 3 i x 4 =– 1 3 i
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