Son Ecuaciones de Segundo Grado o Cuadráticas que se solucionan mediante la Resolvente.
Una Ecuación Cuadrática es de la forma :
Ax² + Bx + C = 0
La Resolvente para hallar las dos Raíces o Soluciones es :
X1, 2 = – B ± √(B² – 4AC) ÷ 2A
Esta se utilizará para resolver todos los problemas dados.
A) x² + 3x – 10 = 0 A = 1 ; B = 3 ; C = – 10
X1, 2 = – (3) ± √[(3)² – 4(1)( – 10)] ÷ 2(1)
X1, 2 = – 3 ± √(9 + 40) ÷ 2
X1, 2 = – 3 ± √49 ÷ 2
X1, 2 = – 3 ± 7 ÷ 2
X1 = – 3 + 7 ÷ 2
X1 = 4 ÷ 2
X1 = 2
X2 = – 3 – 7 ÷ 2
X2 = – 10 ÷ 2
X2 = – 5
b) x² – 3x – 4 = 0 A = 1 ; B = – 3 ; C = – 4
X1, 2 = – (– 3) ± √[(– 3)² – 4(1)( – 4)] ÷ 2(1)
X1, 2 = 3 ± √(9 + 16) ÷ 2
X1, 2 = 3 ± √25 ÷ 2
X1, 2 = 3 ± 5 ÷ 2
X1 = 3 + 5 ÷ 2
X1 = 8 ÷ 2
X1 = 4
X2 = 3 – 5 ÷ 2
X2 = – 2 ÷ 2
X2 = – 1
c) – x² – 4x – 2 = 0 A = – 1 ; B = – 4 ; c = – 2
X1, 2 = – (– 4) ± √[(– 4)² – 4(– 1)( – 2)] ÷ 2(– 1)
X1, 2 = 4 ± √(16 – 8) ÷ – 2
X1, 2 = 4 ± √8 ÷ – 2
X1, 2 = 4 ± 2, 83 ÷ – 2
X1 = 4 + 2, 83 ÷ – 2
X1 = 4 + 2, 83 ÷ – 2
X1 = 6, 83 ÷ – 2
X1 = – 3, 415
X2 = 4 – 2, 83 ÷ – 2
X2 = 1, 17 ÷ – 2
X2 = – 0, 585
d) – 2x² – x = – 6 – 2x² – x + 6 = 0
A = – 2 ; B = – 1 ; C = 6
X1, 2 = – (– 1) ± √[(– 1)² – 4(– 2)(6)] ÷ 2(– 2)
X1, 2 = 1 ± √(1 + 48) ÷ – 4
X1, 2 = 1 ± √49 ÷ – 4
X1, 2 = 1 ± 7 ÷ – 4
X1 = 1 + 7 ÷ – 4
X1 = 8 ÷ – 4
X1 = – 2
X2 = 1 – 7 ÷ – 4
X2 = – 6 ÷ – 4
X2 = – 1.
5
e) (x + 2)² + 1 = 0 x² + 4x + 4 + 1 = 0
x² + 4x + 5 = 0
A = 1 ; B = 4 ; C = 5
X1, 2 = – (4) ± √[(4)² – 4(1)(5)] ÷ 2(1)
X1, 2 = – 4 ± √(16 – 20) ÷ 2
X1, 2 = – 4 ± √(– 4) ÷ 2
No tiene solución en los Reales.
F) (x – 3)² – 4 = 0 x² – 6x + 9 – 4 = 0x² – 6x + 5 = 0
A = 1 ; B = – 6 ; C = 5
X1, 2 = – (– 6) ± √[(– 6)² – 4(1)(5)] ÷ 2(1)
X1, 2 = 6 ± √(36 – 20) ÷ 2
X1, 2 = 6 ± √16 ÷ 2
X1, 2 = 6 ± 4 ÷ 2
X1 = 6 + 4 ÷ 2
X1 = 10 ÷ 2
X1 = 5
X2 = 6 – 4 ÷ 2
X2 = 2 ÷ 2
X2 = 1
g) – 0, 5x² + 2x + 1, 5 = 0
En este caso se puede usar la propiedad de multiplicar ambos lados de la expresión por – 2, quedando :
(– 2)[– 0, 5x² + 2x + 1, 5] = (– 2)0
x² – 4x – 3 = 0
A = 1 ; B = – 4 ; C = – 3
X1, 2 = – (– 4) ± √[(– 4)² – 4(1)( – 3)] ÷ 2(1)
X1, 2 = 4 ± √(16 + 12) ÷ 2
X1, 2 = 4 ± √28 ÷ 2
X1, 2 = 4 ± 5, 29 ÷ 2
X1 = 4 + 5, 29 ÷ 2
X1 = 9, 29 ÷ 2
X1 = 4, 645
X2 = 4 – 5, 29 ÷ 2
X2 = – 1, 29 ÷ 2
X2 = – 0, 645
h) 1, 52 + 2x = 0 2x + 2, 25 = 0 {Ecuación Lineal}
2x = – 2, 25
X = – 2, 25 / 2
X = – 1, 125
i) (x + 2)(x – 3) = 0 x² + 3x + 2x – 6 = 0
x² + 5x – 6 = 0
A = 1 ; B = 5 ; C = – 6
X1, 2 = – (5) ± √[(5)² – 4(1)( – 6)] ÷ 2(1)
Se deja para que el interesado la resuelva y así fije los conocimientos acá indicados.
J) (x + 1)(x – 5) = 16
x² – 5x + x – 5 = 16
x² – 4x – 5 = 16
x² – 4x – 5 – 16 = 0
x² – 4x – 21 = 0
A = 1 ; B = – 4 ; C = – 21
X1, 2 = – (– 4) ± √[(– 4)² – 4(1)( – 21)] ÷ 2(1)
Se deja para que el interesado la resuelva y así fije los conocimientos acá indicados.