Created
October 14, 2019 14:42
-
-
Save spceaza/d07e32ac1a32ecfe44193fa104004396 to your computer and use it in GitHub Desktop.
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
24. | |
x² + 2dx + d² = 0 | |
Se reescribe de otra forma equivalente | |
(1)x² + (2d)x + (d²) = 0 | |
La ecuación tiene la forma: | |
(a)x² + (b )x + (c ) = 0. | |
Donde: | |
a es 1 | |
b es 2d | |
c es d² | |
Se usa la ecuación: | |
x = (-b ± √(b² - 4ac))/2a | |
x = (-(b) ± √((b)² - 4(a)(c))) / 2(a) | |
Se reemplazan los valores | |
x = (-(2d) ± √((2d)² - 4(1)(d²))) / 2(1) | |
x = (-2d ± √(4d² - 4d²)) / 2 | |
x = (-2d ± √(0)) / 2 | |
x = (-2d ± 0) / 2 | |
x = (-2d) / 2 | |
x = -2d / 2 | |
x = -d | |
Solución: x = -d | |
Prueba: | |
Se reemplaza x en x² + 2dx + d² = 0 por el valor encontrado | |
( x)² + 2d( x) + d² = 0 | |
(-d)² + 2d(-d) + d² = 0 | |
d² - 2d(d) + d² = 0 | |
d² - 2d² + d² = 0 | |
d² + d² - 2d² = 0 | |
2d² - 2d² = 0 | |
0 = 0 | |
0 es igual a 0 entonces el resultado es correcto. |
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
26. | |
(x + c)² = d² | |
Se expande la parte izquierda de la igualdad | |
(x + c)² = d² | |
(x + c)(x + c) = d² | |
x² + cx + cx + c² = d² | |
x² + 2cx + c² = d² | |
Se resta d² en ambas partes de la igualdad | |
x² + 2cx + c² = d² | |
x² + 2cx + c² - d² = d² - d² | |
x² + 2cx + c² - d² = 0 | |
Se reescribe de otra forma equivalente | |
(1)x² + (2c)x + (c² - d²) = 0 | |
La ecuación tiene la forma: | |
(a)x² + (b )x + (c ) = 0. | |
Donde: | |
a es 1 | |
b es 2c | |
c es c² - d² | |
Se usa la ecuación: | |
x = (-b ± √(b² - 4ac))/2a | |
x = (-(b) ± √((b)² - 4(a)(c))) / 2(a) | |
Se reemplazan los valores | |
x = (-(2c) ± √((2c)² - 4(1)(c² - d²))) / 2(1) | |
x = (-(2c) ± √((2c)² - 4(c² - d²))) / 2 | |
x = (-2c ± √(4c² - 4(c² - d²))) / 2 | |
x = (-2c ± √(4c² - 4c² + 4d²)) / 2 | |
x = (-2c ± √(0 + 4d²)) / 2 | |
x = (-2c ± √(4d²)) / 2 | |
x = (-2c ± 2√(d²)) / 2 | |
x = (-2c ± 2d) / 2 | |
x = (-2c / 2) ± (2d / 2) | |
x = (-c) ± (d) | |
x = -c ± d | |
Solución: x₁ = -c + d | |
x₂ = -c - d | |
Prueba: | |
Se reemplaza x en (x + c)² = d² por los valores encontrados, es decir: x₁ y x₂ | |
En el caso de x₁ | |
( x₁ + c)² = d² | |
((-c + d) + c)² = d² | |
(-c + d + c)² = d² | |
(d + c - c)² = d² | |
(d)² = d² | |
d² = d² | |
d² es igual a d² entonces el resultado es correcto para x₁ |
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
66. | |
ax² + bx + c = 0 | |
Donde x₁ y x₂ son raices | |
Si x₁ y x₂ son raices, entonces | |
ax₁² + bx₁ + c = 0 | |
ax₂² + bx₂ + c = 0 | |
Para probar que x₁ + x₂ = -b/a | |
Se pueden igualar | |
ax₁² + bx₁ + c = ax₂² + bx₂ + c | |
Se resta c en ambas partes de la igualdad | |
ax₁² + bx₁ + c - c = ax₂² + bx₂ + c - c | |
ax₁² + bx₁ = ax₂² + bx₂ | |
Se resta ax₂² + bx₂ en ambas partes de la igualdad | |
ax₁² + bx₁ - (ax₂² + bx₂) = ax₂² + bx₂ - (ax₂² + bx₂) | |
ax₁² + bx₁ - ax₂² - bx₂ = ax₂² + bx₂ - ax₂² - bx₂ | |
ax₁² + bx₁ - ax₂² - bx₂ = ax₂² - ax₂² - bx₂ + bx₂ | |
ax₁² + bx₁ - ax₂² - bx₂ = 0 | |
ax₁² - ax₂² + bx₁ - bx₂ = 0 | |
a(x₁² - x₂²) + b(x₁ - x₂) = 0 | |
Se resta b(x₁ - x₂) en ambas partes de la igualdad | |
a(x₁² - x₂²) + b(x₁ - x₂) - b(x₁ - x₂) = -b(x₁ - x₂) | |
a(x₁² - x₂²) = -b(x₁ - x₂) | |
Se divide por a en ambas partes de la igualdad | |
a(x₁² - x₂²) / a = -b(x₁ - x₂) / a | |
(x₁² - x₂²) = -b(x₁ - x₂) / a | |
Se divide por (x₁ - x₂) en ambas partes de la igualdad | |
(x₁² - x₂²) / (x₁ - x₂) = -b(x₁ - x₂)/a(x₁ - x₂) | |
(x₁² - x₂²) / (x₁ - x₂) = -b/a | |
Se factoriza (x₁² - x₂²) en (x₁ + x₂)(x₁ - x₂) | |
(x₁ + x₂)(x₁ - x₂) / (x₁ - x₂) = -b/a | |
(x₁ + x₂) = -b/a | |
Se prueba que x₁ + x₂ = -b/a | |
################################################################################################# | |
Para probar que x₁x₂ = c/a | |
Usando la solución anterior | |
x₁ + x₂ = -b/a | |
Se multiplica por -a en ambos lados de la igualdad | |
-a(x₁ + x₂) = -a(-b/a) | |
-a(x₁ + x₂) = ab/a | |
-a(x₁ + x₂) = b | |
Reemplazamos b en ax₁² + bx₁ + c = 0 | |
ax₁² + ( b )x₁ + c = 0 | |
ax₁² + (-a(x₁ + x₂))x₁ + c = 0 | |
ax₁² + (-ax₁ - ax₂)x₁ + c = 0 | |
ax₁² + (-ax₁² - ax₁x₂) + c = 0 | |
ax₁² - ax₁² - ax₁x₂ + c = 0 | |
0 - ax₁x₂ + c = 0 | |
-ax₁x₂ + c = 0 | |
Sumamos ax₁x₂ en ambos lados de la igualdad | |
-ax₁x₂ + c + ax₁x₂ = ax₁x₂ | |
-ax₁x₂ + ax₁x₂ + c = ax₁x₂ | |
0 + c = ax₁x₂ | |
c = ax₁x₂ | |
Dividimos por a en ambos lados de la igualdad | |
c / a = ax₁x₂ / a | |
c / a = x₁x₂ | |
Se prueba que x₁x₂ = c/a |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment