1 / 11

Inecuaciones cuadráticas

CLASE 101. Inecuaciones cuadráticas. 5. 5. –. –. b) x 2 + 3 x + 1 >. b) x 2 + 3 x + 1 >. 4. 4. c) (3 – x )( x + 4) < 3 x +18. c) (3 – x )( x + 4) < 3 x +. Determina el conjunto solución de las siguientes ecuaciones. x 2 – 9. x 2 – 3 x. . a). 3. 6.

brennan-howell
Download Presentation

Inecuaciones cuadráticas

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CLASE 101 Inecuaciones cuadráticas

  2. 5 5 – – b) x2+ 3x + 1 > b) x2+ 3x + 1 > 4 4 c) (3 – x)(x + 4) < 3x +18 c) (3 – x)(x + 4) < 3x + Determina el conjunto solución de las siguientes ecuaciones. x2– 9 x2 – 3x  a) 3 6 d) (x + 1)2  x(4x – 5) – (x – 2)

  3. 3 – 2 3 3 – S ={ x ; x  } 2 2 5 · 4 – b) x2+ 3x + 1 > 4 > – 5 4x + 12x + 4 + 5 > 0 4x + 12x + 4 > 0 4x + 12x + 9 + + (2x + 3)2 > 0 2x + 3 = 0 x = –

  4. c) (3 – x)(x + 4) < 3x + 18 3x + 12 – x2– 4x 3x + 18 > 12 – x2 – 4x – 18 > 0  (–1) – x2 – 4x – 6 > 0 x2 + 4x + 6 > 0 D = 42 – 4  6 D = – 8 0 > S ={ x  }

  5. Para que valores de x el gráfico de la funcion f(x) = (x – 3)2 – 4 está por debajo del gráfico de 4 g(x) = x – 2 3

  6. 4 4 3 3 y g(x) = x – 4 ? ? 1 3 5 0 x (x– 3)2–4 < x – 4 – 4 f(x) =( x– 3)2 – 4

  7. 4 4 3 3 (x– 3)2 – 4 < x – 4 x2 – 6x + 9 < x · 3 3x2 – 18x + 27 <4x 3x2 – 18x + 27 – 4x< 0 3x2 – 22x + 27 < 0

  8. 4 3 D (x– 3)2 – 4 < x – 4 3x2 – 22x + 27 < 0 22 – 12,6 D = 222 – 4  3 27 x1 = 2  3 D = 484 – 324 x1  1,6 D = 160 22 + 12,6 x2 = 2  3  12,6 x2  5,8

  9. 4 4 3 3 y g(x) = x – 4 1,6 ? 5,8 ? + – + 1 3 5 0 x S ={ x ; 1,6< x< 5,8 } (x– 3)2–4 < x – 4 – 4 f(x) =( x– 3)2 – 4

  10. Estudio independiente

  11. LIBRO DE DISTRIBUCIÓN GRATUITA. PROHIBIDA SU VENTA Capítulo 1 Epígrafe 12 Resolver Ejercicio 1

More Related