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Ejercicio 270

Gestor Baldor
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CAPITULO XXXIII

Ecuaciones literales de segundo grado
Ejercicio 270
Resolver las ecuaciones:
  1. x 2 +2ax–35 a 2 =0 ( x+7a ) ( x–5a ) =0 x+7a =0 x 1 =–7a x–5a =0 x 2 =5a
  2. 10 x 2 =36 a 2 –37ax 10 x 2 +37ax–36 a 2 =0 10 x 2 +45ax–8ax–36 a 2 =0 5x( 2x+9a ) –4a( 2x+9a ) =0 ( 2x+9a ) ( 5x–4a ) =0 2x+9a =0 2x =–9a x 1 =– 9a 2 5x–4a =0 5x =4a x 2 = 4a 5
  3. a 2 x 2 +abx–2 b 2 =0 a 2 x 2 –abx+2abx–2 b 2 =0 ax( ax–b ) +2b( ax–b ) =0 ( ax–b ) ( ax+2b ) =0 ax–b =0 ax =b x 1 = b a ax+2b =0 ax =–2b x 2 =– 2b a
  4. 89bx =42 x 2 +22 b 2 42 x 2 –89bx+22 b 2 =0 42 x 2 –12bx–77bx+22 b 2 =0 6x( 7x–2b ) –11b( 7x–2b ) =0 ( 7x–2b ) ( 6x–11b ) =0 7x–2b =0 7x =2b x 1 = 2b 7 6x–11b =0 6x =11b x 2 = 11b 6
  5. x 2 +ax =20 a 2 x 2 +ax–20 a 2 =0 ( x+5a ) ( x–4a ) =0 x+5a =0 x 1 =–5a x–4a =0 x 2 =4a
  6. 2 x 2 =abx+3 a 2 b 2 2 x 2 –abx–3 a 2 b 2 =0 2 x 2 +2abx–3abx–3 a 2 b 2 =0 2x( x+ab ) –3ab( x+ab ) =0 ( x+ab ) ( 2x–3ab ) =0 x+ab =0 x 1 =–ab 2x–3ab =0 2x =3ab x 2 = 3ab 2
  7. b 2 x 2 +2abx =3 a 2 b 2 x 2 +2abx–3 a 2 =0 b 2 x 2 –abx+3abx–3 a 2 =0 bx( bx–a ) +3a( bx–a ) =0 ( bx–a ) ( bx+3a ) =0 bx–a =0 bx =a x 1 = a b bx+3a =0 bx =–3a x 2 =– 3a b
  8. x 2 +ax–bx =ab x 2 +ax–bx–ab =0 x( x+a ) –b( x+a ) =0 ( x+a ) ( x–b ) =0 x+a =0 x 1 =–a x–b =0 x 2 =b
  9. x 2 –2ax =6ab–3bx x 2 –2ax+3bx–6ab =0 x( x–2a ) +3b( x–2a ) =0 ( x–2a ) ( x+3b ) =0 x–2a =0 x 1 =2a x+3b =0 x 2 =–3b
  10. 3( 2 x 2 –mx ) +4nx–2mn =0 3x( 2x–m ) +2n( 2x–m ) =0 ( 3x+2n ) ( 2x–m ) =0 3x+2n =0 3x =–2n x 1 =– 2n 3 2x–m =0 2x =m x 2 = m 2
  11. x 2 – a 2 –bx–ab =0 ( x–a ) ( x+a ) –b( x+a ) =0 ( x+a ) ( x–a–b ) =0 x+a =0 x 1 =–a x–a–b =0 x 2 =a+b
  12. ab x 2 –x( b–2a ) =2 ab x 2 –bx+2ax–2 =0 bx( ax–1 ) +2( ax–1 ) =0 ( ax–1 ) ( bx+2 ) =0 ax–1 =0 ax =1 x 1 = 1 a bx+2 =0 bx =–2 x 2 =– 2 b
  13. x 2 –2ax+ a 2 – b 2 =0 ( x–a ) 2 – b 2 =0 [ ( x–a ) +b ] [ ( x–a ) –b ] =0 ( x–a+b ) ( x–a–b ) =0 x–a+b =0 x 1 =a–b x–a–b =0 x 2 =a+b
  14. 4x( x–b ) + b 2 =4 m 2 4 x 2 –4xb+ b 2 –4 m 2 =0 ( 2x–b ) 2 –4 m 2 =0 [ ( 2x–b ) +2m ] [ ( 2x–b ) –2m ] =0 ( 2x–b+2m ) ( 2x–b–2m ) =0 2x–b–2m =0 2x =b+2m x 1 = b+2m 2 2x–b+2m =0 2x =b–2m x 2 = b–2m 2
  15. x 2 – b 2 +4 a 2 –4ax =0 x 2 –4ax+4 a 2 – b 2 =0 ( x–2a ) 2 – b 2 =0 [ ( x–2a ) +b ] [ ( x–2a ) –b ] =0 ( x–2a+b ) ( x–2a–b ) =0 x–2a–b =0 x 1 =2a+b x–2a+b =0 x 2 =2a–b
  16. x 2 –( a+2 ) x =–2a x 2 –ax–2x+2a =0 x( x–a ) –2( x–a ) =0 ( x–a ) ( x–2 ) =0 x–a =0 x 1 =a x–2 =0 x 1 =2
  17. x 2 +2x( 4–3a ) =48a x 2 +8x–6ax–48a =0 x( x+8 ) –6a( x+8 ) =0 ( x+8 ) ( x–6a ) =0 x+8 =0 x 1 =–8 x–6a =0 x 2 =6a
  18. x 2 –2x = m 2 +2m x 2 – m 2 –2m–2x =0 ( x+m ) ( x–m ) –2( m+x ) =0 ( x+m ) [ ( x–m ) –2 ] =0 ( x+m ) ( x–m–2 ) =0 x+m =0 x 1 =–m x–m–2 =0 x 2 =m+2
  19. x 2 + m 2 x( m–2 ) =2 m 5 x 2 + m 3 x–2 m 2 x–2 m 5 =0 x( x+ m 3 ) –2 m 2 ( x+ m 3 ) =0 ( x–2 m 2 ) ( x+ m 3 ) =0 x+ m 3 =0 x 1 =– m 3 x–2 m 2 =0 x 2 =2 m 2
  20. 6 x 2 –15ax =2bx–5ab 6 x 2 –2bx–15ax+5ab =0 2x( 3x–b ) –5a( 3x–b ) =0 ( 2x–5a ) ( 3x–b ) =0 3x–b =0 3x =b x 1 = b 3 2x–5a =0 2x =5a x 2 = 5a 2
  21. 3x 4 + a 2 – x 2 2a =0 Multiplicando la ecuación por –4a 2 x 2 –3ax–2 a 2 =0 2 x 2 +ax–4ax–2 a 2 =0 x( 2x+a ) –2a( 2x+a ) =0 ( x–2a ) ( 2x+a ) =0 2x+a =0 2x =–a x 1 =– a 2 x–2a =0 x 2 =2a
  22. 2x–b 2 = 2bx– b 2 3x 3x( 2x–b ) =2( 2bx– b 2 ) 6 x 2 –3bx =4bx–2 b 2 6 x 2 –3bx–4bx+2 b 2 =0 6 x 2 –7bx+2 b 2 =0 6 x 2 –3bx–4bx+2 b 2 =0 3x( 2x–b ) –2b( 2x–b ) =0 ( 3x–2b ) ( 2x–b ) =0 2x–b =0 2x =b x 1 = b 2 3x–2b =0 3x =2b x 2 = 2b 3
  23. a+x a–x + a–2x a+x =–4 ( a+x ) 2 +( a–x ) ( a–2x ) ( a–x ) ( a+x ) =–4 a 2 + 2ax + x 2 + a 2 – 2ax –ax+2 x 2 a 2 – x 2 =–4 3 x 2 –ax+2 a 2 =–4( a 2 – x 2 ) 3 x 2 –ax+2 a 2 =–4 a 2 +4 x 2 3 x 2 –ax+2 a 2 +4 a 2 –4 x 2 =0 – x 2 –ax+6 a 2 =0 x 2 +ax–6 a 2 =0 x 2 –2ax+3ax–6 a 2 =0 x( x–2a ) +3a( x–2a ) =0 ( x–2a ) ( x+3a ) =0 x–2a =0 x 1 =2a x+3a =0 x 2 =–3a
  24. x 2 x–1 = a 2 2( a–2 ) 2 x 2 ( a–2 ) = a 2 ( x–1 ) 2a x 2 –4 x 2 = a 2 x– a 2 2a x 2 –4 x 2 – a 2 x+ a 2 =0 2a x 2 – a 2 x+ a 2 –4 x 2 =0 ax( 2x–a ) +( a+2x ) ( a–2x ) =0 –ax( a–2x ) +( a+2x ) ( a–2x ) =0 ( a–2x ) ( –ax+a+2x ) =0 a–2x =0 a =2x x 1 = a 2 –ax+a+2x =0 x( 2–a ) =–a x 2 =– a 2–a ↔ a a–2
  25. x+ 2 x = 1 a +2a x 2 +2 x = 1+2 a 2 a a( x 2 +2 ) =x( 1+2 a 2 ) a x 2 +2a =x+2 a 2 x a x 2 +2a–x–2 a 2 x =0 a x 2 –x–2 a 2 x+2a =0 x( ax–1 ) –2a( ax–1 ) =0 ( x–2a ) ( ax–1 ) =0 x–2a =0 x 1 =2a ax–1 =0 ax =1 x 2 = 1 a
  26. 2x–b b – x x+b = 2x 4b ( 2x–b ) ( x+b ) –bx b ( x+b ) = 2 x b 2 x 2 + 2bx – bx – b 2 – bx x+b = x 2 2( 2 x 2 – b 2 ) =x( x+b ) 4 x 2 –2 b 2 = x 2 +bx 4 x 2 –2 b 2 – x 2 –bx =0 3 x 2 –bx–2 b 2 =0 3 x 2 –3bx+2bx–2 b 2 =0 3x( x–b ) +2b( x–b ) =0 ( 3x+2b ) ( x–b ) =0 x–b =0 x 1 =b 3x+2b =0 3x =–2b x 2 =– 2b 3
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