¿Hola cómo resuelvo esto?
Sea la función f(x)= 3x2+bx+c
Hallar "b" y "c" para que 2 y -1 raíces de f(x).
Arriba lo que quise escribir esa 3x al cuadrado.
Lo que necesito en sí, más que la respuesta, es el paso a paso de cómo resolverlo.
Gracias!!!!!
More Questions From This User See All
Answers & Comments
Por si fuera poco , no agradeciste la primera pregunta.
Creo que ni la revisaste.
https://espanol.answers.yahoo.com/question/index?q...
...
f(x) = ax^2 + bx + c
f(x) = 3x^2 + bx + c
Raíces x1 = 2; x2 = -1
Propiedades de las raíces
Producto
x1*x2 = c/a
2*(-1) = -2
c/a = -2
c = -2*3 = - 6 <========
Suma
x1 + x2 = -b/a
2 + (-1) = 2 - 1 = 1
-b/a = 1
-b = 1*3 = 3
b = -3 <=======
La función original queda
================
f(x) = 3x^2 - 3x - 6 <===========
================
Suerte
numeros!
¿otra vez
3 (x + 1) (x - 2) = 3 (x^2 - x + 2) = 3 x^2 - 3 x + 6
Si no entiendes algo
explica lo que no entiendes
3x² + bx + c = 0
3.[x² + (b/3).x + (c/3)] = 0
x² + (b/3).x + (c/3) = 0
x² + (b/3).x + (b/6)² - (b/6)² + (c/3) = 0
x² + (b/3).x + (b/6)² - (b²/36) + (c/3) = 0
x² + (b/3).x + (b/6)² = (b²/36) - (c/3)
x² + (b/3).x + (b/6)² = (b²/36) - (12c/36)
x² + (b/3).x + (b/6)² = (b² - 12c)/36
[x + (b/6)]² = [± √(b² - 12c)]/6]²
x + (b/6) = ± √(b² - 12c)]/6
x = - (b/6) ± [√(b² - 12c)]/6]
x = [- b ± √(b² - 12c)]/6
First case: x = [- b + √(b² - 12c)]/6 ← you know it's 2
[- b + √(b² - 12c)]/6 = 2
- b + √(b² - 12c) = 12
√(b² - 12c) = 12 + b
{ √(b² - 12c)] }² = (12 + b)²
b² - 12c = (12 + b)²
b² - 12c = 144 + 24b + b²
- 12c = 144 + 24b
- c = 12 + 2b
c = - 12 - 2b ← memorize this result as (1)
Second case: x = [- b - √(b² - 12c)]/6 ← you know it's 1
[- b - √(b² - 12c)]/6 = - 1
- b - √(b² - 12c) = - 6
- √(b² - 12c) = - 6 + b
√(b² - 12c) = 6 - b
{ √(b² - 12c) }² = (6 - b)²
b² - 12c = (6 - b)²
b² - 12c = 36 - 12b + b²
- 12c = 36 - 12b
- c = 3 - b
c = b - 3 → recall (1): c = - 12 - 2b
b - 3 = - 12 - 2b
b + 2b = - 12 + 3
3b = - 9
→ b = - 3
Recall (1): c = - 12 - 2b
c = - 12 + 6
→ c = - 6
Now, let's check the result:
3x² + bx + c = 0 → where: b = - 3 and where: c = - 6
3x² - 3x - 6 = 0
3.(x² - x - 2) = 0
x² - x - 2 = 0
x² - x + (1/2)² - (1/2)² - 2 = 0
x² - x + (1/2)² - (1/4) - (8/4) = 0
x² - x + (1/2)² - (9/4) = 0
x² - x + (1/2)² = 9/4
[x - (1/2)]² = (± 3/2)²
x - (1/2) = ± 3/2
x = (1/2) ± (3/2)
x = (1 ± 3)/2
First case: x = (1 + 3)/2 → x = 4/2 → x = 2
Second case: x = (1 - 3)/2 → x = - 2/2 → x = - 1
SOLUCIÓN:
(SI NO SE VE BIEN, DALE CLICK EN LA IMAGEN:)