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Ejercicio 265

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CAPITULO XXXIII

Ecuaciones de segundo grado
Ejercicio 265
Resolver las siguientes ecuaciones por la fΓ³rmula general:
x= –b Β± b 2 –4ac 2a
  1. 3 x 2 –5x+2 =0 x = 5 Β± 5 2 –4( 3 ) ( 2 ) 2( 3 ) x = 5 Β± 25–24 6 x = 5 Β± 1 6 x 1 = 6 6 x 1 =1 x 2 = x 2 = 2 3
  2. 4 x 2 +3x–22 =0 x = –3 Β± 3 2 –4( 4 ) ( –22 ) 2( 4 ) x = –3 Β± 9–352 8 x = –3 Β± 361 8 x = –3 Β± 19 8 x 1 = 8 x 1 =2 x 2 = – x 2 =– 11 4
  3. x 2 +11x =–24 x 2 +11x+24 =0 x = –11 Β± 1 1 2 –4( 1 ) ( 24 ) 2( 1 ) x = –11 Β± 121–96 2 x = –11 Β± 25 2 x = –11 Β± 5 2 x 1 = – 2 x 1 =–8 x 2 = – 2 x 2 =–3
  4. x 2 =16x–63 x 2 –16x+63 =0 x = 16 Β± 1 6 2 –4( 1 ) ( 63 ) 2( 1 ) x = 16 Β± 1 6 2 –4( 1 ) ( 63 ) 2( 1 ) x = 16 Β± 256–252 2 x = 16 Β± 4 2 x = 16 Β± 2 2 x 1 = 2 x 1 =9 x 2 = 2 x 2 =7
  5. 12x–4–9 x 2 =0 9 x 2 –12x+4 =0 x = 12 Β± 1 2 2 –4( 9 ) ( 4 ) 2( 9 ) x = 12 Β± 18 x = x = 2 3
  6. 5 x 2 –7x–90 =0 x = 7 Β± 7 2 –4( 5 ) ( –90 ) 2( 5 ) x = 7 Β± 49+1800 10 x = 7 Β± 1849 10 x = 7 Β± 43 10 x 1 = 5 0 1 0 x 1 =5 x 2 = – x 2 =– 18 5
  7. 6 x 2 =x+222 6 x 2 –x–222 =0 x = 1 Β± 1–4( 6 ) ( –222 ) 2( 6 ) x = 1 Β± 1+5328 12 x = 1 Β± 73 12 x 1 = x 1 = 37 6 x 2 = – 12 x 2 =–6
  8. x+11 =10 x 2 10 x 2 –x–11 =0 x = 1 Β± 1–4( 10 ) ( –11 ) 2( 10 ) x = 1 Β± 1+440 20 x = 1 Β± 21 20 x 1 = x 1 = 11 10 x 2 = – 20 20 x 2 =–1
  9. 49 x 2 –70x+25 =0 x = 70 Β± 7 0 2 –4( 49 ) ( 25 ) 2( 49 ) x = 70 Β± 98 x = x = 5 7
  10. 12x–7 x 2 +64 =0 7 x 2 –12x–64 =0 x = 12 Β± 1 2 2 –4( 7 ) ( –64 ) 2( 7 ) x = 12 Β± 144+1792 14 x = 12 Β± 1936 14 x = 12 Β± 44 14 x 1 = 14 x 1 =4 x 2 = – x 2 =– 16 7
  11. x 2 =–15x–56 x 2 +15x+56 =0 x = –15 Β± 1 5 2 –4( 1 ) ( 56 ) 2( 1 ) x = –15 Β± 225–224 2 x = –15 Β± 1 2 x 1 = – 2 x 1 =–7 x 2 = – 2 x 2 =–8
  12. 32 x 2 +18x–17 =0 x = –18 Β± 1 8 2 –4( 32 ) ( –17 ) 2( 32 ) x = –18 Β± 324+2176 64 x = –18 Β± 2500 64 x = –18 Β± 50 64 x 1 = 32 x 1 = 1 2 x 2 = – x 2 =– 17 16
  13. 176x =121+64 x 2 64 x 2 –176x+121 =0 x = –176 Β± 17 6 2 –4( 64 ) ( 121 ) 2( 64 ) x = –176 Β± 128 x =– x =– 11 8
  14. 8x+5 =36 x 2 36 x 2 –8x–5 =0 x = 8 Β± 8 2 –4( 36 ) ( –5 ) 2( 36 ) x = 8 Β± 64+720 72 x = 8 Β± 784 72 x = 8 Β± 28 72 x 1 = 36 x 1 = 1 2 x 2 =– x 2 =– 5 18
  15. 27 x 2 +12x–7 =0 x = –12 Β± 1 2 2 –4( 27 ) ( –7 ) 2( 27 ) x = –12 Β± 144+756 54 x = –12 Β± 900 54 x = –12 Β± 30 54 x 1 = 18 x 1 = 1 3 x 2 =– x 2 =– 7 9
  16. 15x =25 x 2 +2 25 x 2 –15x+2 =0 x = 15 Β± 1 5 2 –4( 25 ) ( 2 ) 2( 25 ) x = 15 Β± 225–200 50 x = 15 Β± 25 50 x = 15 Β± 5 50 x 1 = 2 0 5 0 x 1 = 2 5 x 2 = 1 0 5 0 x 2 = 1 5
  17. 8 x 2 –2x–3 =0 x = 2 Β± 2 2 –4( 8 ) ( –3 ) 2( 8 ) x = 2 Β± 4+96 16 x = 2 Β± 100 16 x = 2 Β± 10 16 x 1 = x 1 = 3 4 x 2 = – 8 x 2 =– 1 2
  18. 105 =x+2 x 2 2 x 2 +x–105 =0 x = –1 Β± 1–4( 2 ) ( –105 ) 2( 2 ) x = –1 Β± 1–4( 2 ) ( –105 ) 4 x = –1 Β± 1+840 4 x = –1 Β± 841 4 x = –1 Β± 29 4 x 1 = 4 x 1 =7 x 2 = – x 2 =– 15 2
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