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Ejercicio 266

Gestor Baldor
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CAPITULO XXXIII

Ecuaciones de segundo grado
Ejercicio 266
Resolver las siguientes ecuaciones llevándolas a la forma a x 2 +bx+c=0 y aplicando la fórmula general:
x= –b ± b 2 –4ac 2a
  1. x( x+3 ) =5x+3 x 2 +3x =5x+3 x 2 +3x–5x–3 =0 x 2 –2x–3 =0 x = 2 ± 2 2 –4( 1 ) ( –3 ) 2( 1 ) x = 2 ± 4+12 2 x = 2 ± 16 2 x = 2 ± 4 2 x 1 = 2 x 1 =3 x 2 = – 2 2 x 2 =–1
  2. 3( 3x–2 ) =( x+4 ) ( 4–x ) 9x–6 = 4x – x 2 +16– 4x 9x–6–16+ x 2 =0 x 2 +9x–22 =0 x = –9 ± 9 2 –4( 1 ) ( –22 ) 2( 1 ) x = –9 ± 81+88 2 x = –9 ± 169 2 x = –3 ± 13 2 x 1 = 2 x 1 =5 x 2 = – 2 x 2 =–4
  3. 9x+1 =3( x 2 –5 ) –( x–3 ) ( x+2 ) 9x+1 =3 x 2 –15–( x 2 –x–6 ) 9x+1 =3 x 2 –15– x 2 +x+6 0 =2 x 2 –9+x–9x–1 0 =2 x 2 –8x–10 Dividiendo la ecuación para 2 x 2 –4x–5 =0 x = –4 ± 4 2 –4( 1 ) ( –5 ) 2( 1 ) x = –4 ± 16+20 2 x = –4 ± 36 2 x = –4 ± 6 2 x 1 = 2 2 x 1 =1 x 2 = – 2 x 2 =–5
  4. ( 2x–3 ) 2 – ( x+5 ) 2 =–23 4 x 2 –12x+9–( x 2 +10x+25 ) =–23 4 x 2 –12x+9– x 2 –10x–25 =–23 3 x 2 –22x–16+23 =0 3 x 2 –22x+7 =0 x = 22 ± 2 2 2 –4( 3 ) ( 7 ) 2( 3 ) x = 22 ± 484–84 6 x = 22 ± 400 6 x = 22 ± 20 6 x 1 = 2 x 1 = 1 3 x 2 = 6 x 2 =7
  5. 25 ( x+2 ) 2 = ( x–7 ) 2 –81 25( x 2 +4x+4 ) = x 2 –14x+49–81 25 x 2 +100x+100 = x 2 –14x–32 25 x 2 +100x+100– x 2 +14x+32 =0 24 x 2 +114x+132 =0 Dividiendo la ecuación para 6 4 x 2 +19x+22 =0 x = –19 ± 1 9 2 –4( 4 ) ( 22 ) 2( 4 ) x = –19 ± 361–352 8 x = –19 ± 9 8 x = –19 ± 3 8 x 1 = – x 1 =– 11 4 x 2 = – 8 x 2 =–2
  6. 3x( x–2 ) –( x–6 ) =23( x–3 ) 3 x 2 –6x–x+6 =23x–69 3 x 2 –7x+6–23x+69 =0 3 x 2 –30x+75 =0 Dividiendo la ecuación para 3 x 2 –10x+25 =0 x = 10 ± 1 0 2 –4( 1 ) ( 25 ) 2( 1 ) x = 10 ± 2 x = 2 x =5
  7. 7( x–3 ) –5( x 2 –1 ) = x 2 –5( x+2 ) 7x–21–5 x 2 +5 = x 2 –5x–10 7x–16–5 x 2 – x 2 +5x+10 =0 –6 x 2 +12x–6 =0 Dividiendo para –6 la ecuación x 2 –2x+1 =0 x = 2 ± 2 2 –4( 1 ) ( 1 ) 2( 1 ) x = 2 ± 2 x = 2 2 x =1
  8. ( x–5 ) 2 – ( x–6 ) 2 = ( 2x–3 ) 2 –118 x 2 –10x+25–( x 2 –12x+36 ) =4 x 2 –12x+9–118 x 2 –10x+25– x 2 +12x–36 =4 x 2 –12x–109 0 =4 x 2 –12x–109–2x+11 0 =4 x 2 –14x–98 Dividimos la ecuación para 2 2 x 2 –7x–49 =0 x = 7 ± 7 2 –4( 2 ) ( –49 ) 2( 2 ) x = 7 ± 49+392 4 x = 7 ± 441 4 x = 7 ± 21 4 x 1 = 4 x 1 =7 x 2 = – x 2 =– 7 2
  9. ( 5x–2 ) 2 – ( 3x+1 ) 2 – x 2 –60 =0 25 x 2 –20x+4–( 9 x 2 +6x+1 ) – x 2 –60 =0 25 x 2 –20x–9 x 2 –6x–1– x 2 –56 =0 15 x 2 –26x–57 =0 x = 26 ± 2 6 2 –4( 15 ) ( –57 ) 2( 15 ) x = 26 ± 676+3420 30 x = 26 ± 4096 30 x = 26 ± 64 30 x 1 = 30 x 1 =3 x 2 = – x 2 =– 19 15
  10. ( x+4 ) 3 – ( x–3 ) 3 =343 x 3 +12 x 2 +48x+64–( x 3 –9 x 2 +27x–27 ) =343 x 3 +12 x 2 +48x+64– x 3 +9 x 2 –27x+27–343 =0 21 x 2 +21x–252 =0 Dividiendo la ecuación para 21 x 2 +x–12 =0 x = –1 ± 1–4( 1 ) ( –12 ) 2( 1 ) x = –1 ± 1+48 2 x = –1 ± 49 2 x = –1 ± 7 2 x 1 = – 2 x 1 =–4 x 2 = 2 x 2 =3
  11. ( x+2 ) 3 – ( x–1 ) 3 =x( 3x+4 ) +8 x 3 +6 x 2 +12x+8–( x 3 –3 x 2 +3x–1 ) =3 x 2 +4x+8 x 3 +6 x 2 +12x+ 8 – x 3 + 3 x 2 –3x+1 = 3 x 2 +4x+ 8 6 x 2 +9x+1–4x =0 6 x 2 +5x+1 =0 x = –5 ± 5 2 –4( 6 ) ( 1 ) 2( 6 ) x = –5 ± 12 x = –5 ± 1 12 x 1 = – 4 x 1 =– 1 3 x 2 = – 6 x 2 =– 1 2
  12. ( 5x–4 ) 2 –( 3x+5 ) ( 2x–1 ) =20x( x–2 ) +27 25 x 2 – 40x +16–( 6 x 2 –3x+10x–5 ) =20 x 2 – 40x +27 25 x 2 +16–6 x 2 –7x+5–20 x 2 –27 =0 – x 2 –7x–6 =0 x 2 +7x+6 =0 x = –7 ± 7 2 –4( 1 ) ( 6 ) 2( 1 ) x = –7 ± 49–24 2 x = –7 ± 25 2 x = –7 ± 5 2 x 1 = – 2 2 x 1 =–1 x 2 = – 2 x 2 =–6
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