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3. C R E E M O S E N L A E X I G E N C I A
Las ecuaciones nos
sirven para calcular
valores desconocidos,
de ahí la palabra
“incógnita”
C U R S O D E Á L G E B R A
Las ecuaciones son conocidas de
hace mucho tiempo, ya aparecían
en el Papiro de Rhind del siglo XVI
a.n.e
"Un montón y un séptimo del
mismo es igual a 24".
Gracias al desarrollo de las
ecuaciones se ha logrado el
avance en la arquitectura,
escultura, mecánica y las
demás ciencias, siendo una
parte fundamental en el
desarrollo tecnológico
4. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
OBJETIVOS
✓ Aplicar las propiedades de las
ecuaciones cuadráticas.
✓ Resolver ecuaciones de primer
grado
✓ Resolver ecuaciones de segundo
grado.
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
Arco parabólico Tacna
LAS MATEMÁTICAS EN LA ARQUITECTURA
El Arco Parabólico se encuentra en el
Paseo Cívico de la ciudad de Tacna,
en honor a los héroes del pacífico:
Miguel Grau y Francisco Bolognesi.
Tiene una altura de 18 m
y forma parabólica.
5. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
ECUACIONES
Una ecuación es una igualdad de dos expresiones
matemáticas, donde existe al menos una variable o
incógnita.
Ejemplos:
I) 𝑥2
= 25
II) 𝑥 − 1 0
= 1
III)
1
𝑥 − 2
= 0
Solución
Es el valor de la incógnita que verifica la igualdad.
Ejemplos:
I)En la ecuación 𝑥2 = 25, tenemos que 𝑥 = 5 es solución
porque:
52= 25
II)En 𝑥 − 1 0
= 1, tenemos que 𝑥 = 8 es solución
porque:
8 − 1 0
= 1
III) La ecuación
1
𝑥 − 2
= 0 no tiene soluciones.
6. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Conjunto solución (C.S)
Es el conjunto conformado por todas las soluciones
de una ecuación
Ejemplos:
I) 𝑥2
= 25
La ecuación tiene como
conjunto solución 𝐶. 𝑆 = −5; 5
II) 𝑥 − 1 0 = 1
La ecuación tiene como
conjunto solución 𝐶. 𝑆 = ℝ − 1
III)
1
𝑥 − 2
= 0
La ecuación tiene como
conjunto solución 𝐶. 𝑆 = 𝜙
Clases de ecuaciones
Por su conjunto solución
1) Ecuación compatible: Tiene al menos una solución
I) Si el número de soluciones es finito, se denomina
compatible determinado.
𝑥2
= 25 𝐶. 𝑆 = −5; 5
II) Si el número de soluciones es infinito, se denomina
compatible indeterminado.
𝑥 − 1 0
= 1 𝐶. 𝑆 = ℝ − 1
2) Ecuación incompatible: No tiene solución.
1
𝑥 − 2
= 0 𝐶. 𝑆 = 𝜙
8. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ecuación polinomial
Es aquella ecuación que presenta la siguiente forma
general
𝑎0𝑥𝑛
+ 𝑎1𝑥𝑛−1
+ 𝑎2𝑥𝑛−2
+ ⋯ 𝑎𝑛−1𝑥 + 𝑎𝑛 = 0
donde:
𝑎0; 𝑎1; 𝑎2; … ; 𝑎𝑛 son los coeficientes (𝑎0 ≠ 0)
𝑥 es la incógnita
El grado del polinomio determina el grado
de la ecuación
NOTA:
Ejemplos
• 3𝑥 + 2 = 0 Ecuación lineal o de primer grado
• 5𝑥2 − 8𝑥 − 3 = 0 Ecuación cuadrática
• 2𝑥3 + 9𝑥 − 5 = 0 Ecuación cúbica
I) Ecuación lineal
Llamadas ecuaciones polinomiales de primer grado.
Su forma general es: 𝐴𝑥 + 𝐵 = 0 ; 𝐴 ≠ 0
Resolución
Despejamos las variables en un miembro y en el otro
miembro las constantes; de ahí se encuentra el valor
de la variable.
Ejemplo
3𝑥 + 2 = 0 3𝑥 = −2 𝑥 = −
2
3
∴ 𝐶. 𝑆 = −
2
3
NOTA: Resolver una ecuación es encontrar su C.S
9. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
II) Ecuación cuadrática
Llamadas ecuaciones polinomiales de segundo grado.
Su forma general es:
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 ; 𝑎 ≠ 0
Resolución
1) Por factorización:
a) Se factoriza el polinomio cuadrático.
b) Se utiliza el siguiente teorema.
𝑎𝑏 = 0 ↔ 𝑎 = 0 ∨ 𝑏 = 0
c) Se encuentran las soluciones de la ecuación.
d) Se encuentra el conjunto solución.
Ejemplo 1
Resolver :
3𝑥2 + 𝑥 − 10 = 0
Resolución
Factorizando
3𝑥
𝑥
−5
+2
3𝑥 − 5 𝑥 + 2 = 0
3𝑥 − 5 = 0 ∨ 𝑥 + 2 = 0
𝑥 =
5
3
∨ 𝑥 = −2
∴ 𝐶. 𝑆 =
5
3
; −2
3𝑥2
+ 𝑥 − 10 = 0
10. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ejemplo 2
Resolver :
Resolución
4𝑥2
− 28𝑥 + 49 = 0
4𝑥2 − 28𝑥 + 49 = 0
Factorizando
2𝑥
2𝑥
−7
−7
2𝑥 − 7 2𝑥 − 7 = 0
2𝑥 − 7 = 0 ∨ 2𝑥 − 7 = 0
𝑥 =
7
2
∨
∴ 𝐶. 𝑆 =
7
2
𝑥 =
7
2
Ejemplo
Resolver :
Resolución
4𝑥2 − 4𝑥 − 5 = 0
4𝑥2
− 4𝑥 = 5
Despejando :
Se forma T.C.P
+1 +1
2𝑥 − 1 2
Queda: 2𝑥 − 1 2 = 6
Teorema: 𝑥2
= 𝑎 → 𝑥 = 𝑎 ∨ 𝑥 = − 𝑎
2𝑥 − 1 = 6 ∨ 2𝑥 − 1 = − 6
𝑥 =
1 + 6
2
∨ 𝑥 =
1 − 6
2
∴ 𝐶. 𝑆 =
1 + 6
2
;
1 − 6
2
2) Por completación de cuadrados:
11. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
3) Por fórmula general:
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 ; 𝑎 ≠ 0
Tenemos:
÷ 𝑎
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
𝑎 𝑎 𝑎 𝑎
𝑥2
+
𝑏
𝑎
𝑥 +
𝑐
𝑎
= 0
𝑥2
+
𝑏
𝑎
𝑥 = −
𝑐
𝑎
Se busca un T.C.P.
+
𝑏
2𝑎
2
𝑏
2𝑎
2
𝑥2 +
𝑏
𝑎
𝑥 = −
𝑐
𝑎
𝑥 +
𝑏
2𝑎
2
=
𝑏2
4𝑎2 −
𝑐
𝑎
𝑥 +
𝑏
2𝑎
2
=
𝑏2
− 4𝑎𝑐
4𝑎2
×
4𝑎
4𝑎
Teorema: 𝑥2 = 𝑎 → 𝑥 = ± 𝑎
𝑥 +
𝑏
2𝑎
= ±
𝑏2 − 4𝑎𝑐
4𝑎2 = ±
𝑏2 − 4𝑎𝑐
2𝑎
𝑥 +
𝑏
2𝑎
= ±
𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
12. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ejemplo 1
Resolver : 2𝑥2
+ 4𝑥 − 3 = 0
Resolución
Tenemos
2𝑥2
+ 4 𝑥 − 3 = 0
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
Como: 𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−(4) ± (4)2−4 2 −3
2(2)
𝑥 =
−4 ± 16 + 24
4
=
−4 ± 40
4
𝑥 =
−2 ± 10
2
Luego:
𝑥1 =
−2 + 10
2
∨ 𝑥2 =
−2 − 10
2
Entonces
𝐶. 𝑆 =
−2 + 10
2
;
−2 − 10
2
=
−4 ± 2 10
4
2 1
2
40 = 4 10
= 4 10 = 2 10
13. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ejemplo 2
Resolver : 3𝑥2
+ 𝑥 − 10 = 0
Resolución
Tenemos
3𝑥2
+ 𝑥 − 10 = 0
1
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
Como: 𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−(1) ± (1)2−4 3 −10
2(3)
𝑥 =
−1 ± 1 + 120
6
=
−1 ± 121
6
𝑥 =
−1 ± 11
6
Luego:
𝑥1 =
−1 + 11
6
∨ 𝑥2 =
−1 − 11
6
𝑥1 =
10
6
∨ 𝑥2 =
−12
6
=
5
3
= −2
Entonces
𝐶. 𝑆 =
5
3
; −2
14. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ejemplo 3
Resolver : 9𝑥2
+ 12𝑥 + 4 = 0
Resolución
Tenemos
9𝑥2
+ 12𝑥 + 4 = 0
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
Como: 𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−(12) ± (12)2−4 9 4
2(9)
𝑥 =
−12 ± 144 − 144
18
=
−12 ± 0
18
𝑥 =
−12 ± 0
18
Luego:
𝑥1 =
−12 + 0
18
∨ 𝑥2 =
−12 − 0
18
𝑥1 =
−12
18
∨ 𝑥2 =
−12
18
=
−2
3
Entonces
𝐶. 𝑆 =
−2
3
=
−2
3
15. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ejemplo 4
Resolver : 𝑥2
− 6𝑥 + 10 = 0
Resolución
Tenemos
𝑥2
− 6 𝑥 + 10 = 0
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
Como: 𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−(−6) ± (−6)2−4 1 10
2(1)
𝑥 =
6 ± 36 − 40
2
=
6 ± −4
2
𝑥 =
6 ± 2𝑖
2
Luego:
𝑥1 =
6 + 2𝑖
2
∨ 𝑥2 =
6 − 2𝑖
2
𝑥1 = 3 + 𝑖 ∨ 𝑥2 = 3 − 𝑖
Entonces
𝐶. 𝑆 = 3 + 𝑖; 3 − 𝑖
1
−4 = 4 −1
= 4 −1 = 2𝑖
16. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Resumen:
Las raíces de 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 están dadas
por:
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
Esto es:
𝑥1 =
−𝑏 + ∆
2𝑎
∨ 𝑥2 =
−𝑏 − ∆
2𝑎
Donde:
∆ = 𝑏2 − 4𝑎𝑐
A ∆ se le denomina discriminante
NOTA:
Si los coeficientes de la ecuación 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
son números reales, entonces:
I) Si ∆ > 0: La ecuación presenta dos raíces reales y
diferentes.
𝐶. 𝑆 = 𝑥1; 𝑥2 ; 𝑥1; 𝑥2 ⊂ ℝ
II) Si ∆= 0: La ecuación presenta dos raíces reales e
iguales.
𝑥1 = 𝑥2 = 𝛼 𝐶. 𝑆 = 𝛼
III) Si ∆ < 0: La ecuación presenta dos raíces no
reales (complejas conjugadas).
𝑥1 = 𝑚 + 𝑛𝑖 𝑥2 = 𝑚 − 𝑛𝑖
;
17. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
II.3) Teorema de Cardano - Viette:
Si 𝑥1; 𝑥2 son las raíces de la ecuación:
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 ; 𝑎 ≠ 0
Se cumple:
𝑥1 + 𝑥2 = −
𝑏
𝑎
Suma de raíces
𝑥1. 𝑥2 =
𝑐
𝑎
Producto de raíces
NOTA: Para calcular la diferencia de raíces.
𝑥1 + 𝑥2
2
− 𝑥1 − 𝑥2
2
= 4𝑥1. 𝑥2
Ejemplo:
Si 𝛼; 𝛽 son las raíces de la ecuación 𝑥2
+ 𝑝𝑥 + 36 = 0
tal que:
1
𝛼
+
1
𝛽
=
5
12
. Halle el valor de p.
Resolución:
Por el teorema de Cardano – Viette, tenemos:
𝛼 + 𝛽 = −
𝑏
𝑎
= −
𝑝
1
𝛼 + 𝛽 = −p
𝛼. 𝛽 =
𝑐
𝑎 =
36
1
𝛼. 𝛽 = 36
Como:
1
𝛼
+
1
𝛽
=
𝛼 + 𝛽
𝛼. 𝛽
5
12
=
−𝑝
36
𝑝 = −15
18. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Raíces simétricas: Decimos que 𝑥1; 𝑥2 son raíces
simétricas si
𝑥1 + 𝑥2 = 0
Teorema: Si una ecuación cuadrática
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 ; 𝑎 ≠ 0
tiene raíces simétricas, entonces 𝑏 = 0
Ejemplo:
Si la ecuación 2𝑥2
+ (𝑎 − 2)𝑥 + (𝑎 + 4) = 0
tiene raíces simétricas, calcule 𝑎
Como la ecuación tiene raíces simétricas, por
teorema:
𝑎 − 2 = 0 𝑎 = 2
Raíces recíprocas: Decimos que 𝑥1; 𝑥2 son raíces
simétricas si
𝑥1. 𝑥2 = 1
Teorema: Si una ecuación cuadrática
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 ; 𝑎 ≠ 0
tiene raíces simétricas, entonces 𝑎 = 𝑐
Ejemplo:
Si la ecuación (2𝑛 + 3)𝑥2
+ (3𝑛 − 4)𝑥 + (𝑛 + 8) = 0
tiene raíces simétricas, calcule 𝑛
Como la ecuación tiene raíces recíprocas, por teorema:
2𝑛 + 3 = 𝑛 + 8 𝑛 = 5
19. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
II.4) Reconstrucción de ecuación cuadrática
La ecuación cuadrática cuyas raíces son 𝑥1; 𝑥2 es:
𝑥 − 𝑥1 𝑥 − 𝑥2 = 0
𝑥2
− 𝑥1 + 𝑥2 𝑥 + 𝑥1𝑥2 = 0
S P
Nos queda:
𝑥2
− 𝑆𝑥 + 𝑃 = 0
Donde:
𝑆 = 𝑥1 + 𝑥2
𝑃 = 𝑥1. 𝑥2
Suma de raíces
Producto de raíces
Ejemplo:
Encuentre la ecuación cuadrática cuyas raíces son
𝑥1 = 1 + 5 ; 𝑥2 = 1 − 5
Resolución:
𝑆 = 𝑥1 + 𝑥2 = 1 + 5 + 1 − 5 = 2
𝑃 = 𝑥1. 𝑥2 = 1 + 5 1 − 5 = 12
− 5
2
= −4
Tenemos:
La ecuación cuadrática pedida es:
𝑥2
− 𝑆𝑥 + 𝑃 = 0
𝑥2
− 2𝑥 + (−4) = 0
𝑥2
− 2𝑥 − 4 = 0
20. C R E E M O S E N L A E X I G E N C I A
II.4) Ecuaciones equivalentes
C U R S O D E Á L G E B R A
Se llaman ecuaciones equivalentes, si sus conjuntos
soluciones son iguales.
Ejemplo:
2𝑥 − 10 = 0
Dadas las ecuaciones:
𝑥2
− 10𝑥 + 25 = 0
Tenemos que:
2𝑥 − 10 = 0 𝐶. 𝑆 = 5
𝑥2
− 10𝑥 + 25 = 0 𝑥 − 5 2
= 0
𝑥 − 5 = 0 𝐶. 𝑆 = 5
La ecuaciones son equivalentes
Teorema: Si las ecuaciones
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
𝑚𝑥2 + 𝑛𝑥 + 𝑝 = 0
; 𝑎. 𝑏. 𝑐 ≠ 0
; 𝑚. 𝑛. 𝑝 ≠ 0
Son equivalentes, entonces
𝑎
𝑚
=
𝑏
𝑛
=
𝑐
𝑝
Ejemplo:
Si las ecuaciones:
𝑥2 + 𝑏𝑥 + 3 = 0 ∧ 2𝑥2
+ 8𝑥 + 𝑐 = 0
Son equivalentes, calcule 𝑏. 𝑐
Por teorema:
1
2
=
𝑏
8
=
3
𝑐
𝑏 = 4
𝑐 = 6
𝑏. 𝑐 = 24
Tienen el
mismo C.S
21. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e