Icono del sitio Solucionario Baldor

Ejercicio 278

Comparte esto πŸ‘πŸ‘
CAPITULO XXXV

Dadas las raΓ­ces de una ecuaciΓ³n de segundo grado, determinar la ecuaciΓ³n
Ejercicio 278
Determinar la ecuaciΓ³n cuyas raΓ­ces son:
  1. 3 y 4
    3+4 =7 ∧ β‡’ x 2 –7x+12=0 3( 4 ) =12
  2. -1 y 3
    –1+3 =2 ∧ β‡’ x 2 –2x–3=0 –1( 3 ) =–3
  3. -5 y -7
    –5–7 =–12 ∧ β‡’ x 2 +12x+35=0 –5( –7 ) =35
  4. -10 y 11
    –10+11 =1 ∧ β‡’ x 2 –x+110=0 –10( 11 ) =110
  5. 1 y 1 2
    1+ 1 2 = 3 2 ∧ β‡’ x 2 – 3 2 x+ 1 2 =0 ↔ 2 x 2 –3x+1=0 1( 1 2 ) = 1 2
  6. -2 y – 1 5
    –2– 1 5 =– 11 5 ∧ β‡’ x 2 + 11 5 x+ 2 5 =0 ↔ 5 x 2 +11x+2=0 –2( – 1 5 ) = 2 5
  7. 3 y – 2 3
    3– 2 3 = 7 3 ∧ β‡’ x 2 – 7 3 x+1=0 ↔ 3 x 2 –7x+3=0 3 ( 1 3 ) =1
  8. -2 y – 3 2
    –2– 3 2 =– 7 2 ∧ β‡’ x 2 + 7 2 x+3=0 ↔ 2 x 2 +7x+6=0 – 2 ( – 3 2 ) =3
  9. – 1 2 y 3 4
    – 1 2 + 3 4 = 1 4 ∧ β‡’ x 2 – 1 4 x– 3 8 =0 ↔ 8 x 2 –2x–3=0 – 1 2 ( 3 4 ) =– 3 8
  10. -5 y 2 7
    –5+ 2 7 =– 33 7 ∧ β‡’ x 2 + 33 7 x– 10 7 =0 ↔ 7 x 2 +33x–10=0 –5( 2 7 ) =– 10 7
  11. 6 y – 5 3
    6– 5 3 = 13 3 ∧ β‡’ x 2 – 13 3 x–10=0 ↔ 3 x 2 –13x–30=0 ( – 5 3 ) =–10
  12. -2 y – 1 8
    –2– 1 8 =– 17 8 ∧ β‡’ x 2 + 17 8 x+ 1 4 =0 ↔ 8 x 2 +17x+2=0 – 2 ( – 1 ) = 1 4
  13. 18 y -52
    18–52 =–34 ∧ β‡’ x 2 +34x+936=0 18( –52 ) =936
  14. -15 y -11
    –15–11 =–26 ∧ β‡’ x 2 +26x+165=0 –15( –11 ) =165
  15. 0 y 2
    0+2 =2 ∧ β‡’ x 2 –2x=0 0( 2 ) =0
  16. 0 y – 1 3
    0– 1 3 =– 1 3 ∧ β‡’ x 2 + 1 3 x=0 ↔ 3 x 2 +x=0 0( – 1 3 ) =0
  17. 5 y -5
    5–5 =0 ∧ β‡’ x 2 –25=0 5( –5 ) =–25
  18. 1 2 y – 1 2
    1 2 – 1 2 =0 ∧ β‡’ x 2 – 1 4 =0 ↔ 4 x 2 –1=0 1 2 ( – 1 2 ) =– 1 4
  19. 7 y 7
    7+7 =14 ∧ β‡’ x 2 –14x+49=0 7( 7 ) =49
  20. 8 y – 11 3
    8– 11 3 = 13 3 ∧ β‡’ x 2 – 13 3 x– 88 3 =0 ↔ 3 x 2 –13x–88=0 8( – 11 3 ) =– 88 3
  21. – 5 6 y – 9 2
    – 5 6 – 9 2 =– 16 3 ∧ β‡’ x 2 + 16 3 x+ 15 4 =0 ↔ 12 x 2 +64x+45=0 – 5 ( – 2 ) = 15 4
  22. – 11 2 y 2 7
    – 11 2 + 2 7 =– 73 14 ∧ β‡’ x 2 + 73 14 x– 11 7 =0 ↔ 14 x 2 +73x–22=0 – 11 2 ( 2 7 ) =– 11 7
  23. 2a y -a
    2a–a =a ∧ β‡’ x 2 –ax–2 a 2 =0 2a( –a ) =–2 a 2
  24. – 2b 3 y b 4
    – 2b 3 + b 4 =– 5 12 b ∧ β‡’ x 2 + 5 12 bx– b 2 6 =0 ↔ 12 x 2 +5bx–2 b 2 =0 – 2 b 3 ( b ) =– b 2 6
  25. m y – m 2
    m– m 2 = m 2 ∧ β‡’ x 2 – m 2 x– m 2 2 =0 ↔ 2 x 2 –mx– m 2 =0 m( – m 2 ) =– m 2 2
Salir de la versiΓ³n mΓ³vil