-×^2-4×-2=0
(×-3)^2-4=0
-2x^2-x=-6
-0,5x^2+2×+1,5=0
(X+2)^2+1=0
1,5×^2+2x=0
Respuestas a la pregunta
Respuesta:
Explicación paso a paso:
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 = 0
x² – 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)
Me salio.