Pregunta 1. Resuelva las siguientes ecuaciones cuadráticas:
i) x 2 + 10ix – 21 = 0
Solución:
x2 + 10ix – 21 = 0
x2 + 7ix + 3ix – 21 = 0
x(x + 7i) + 3i(x + 7i) = 0
(x + 7i)(x + 3i) = 0
x + 7i = 0 ; x + 3i = 0;
x = -7i, -3i
Por lo tanto, las raíces son -7i, -3i
ii)x2 + (1 – 2x)x – 2i = 0
Solución:
x 2 + x – 2ix – 2i = 0
x(x+1) – 2i(x+1) = 0
(x + 1)(x – 2i) = 0
x = 2i, -1
iii) x2 – (2√3 + 3i)x + 6√3i = 0
Solución:
x 2 – 2√3x – 3ix + 6√3i = 0
x(x – 2√3) – 3i(x – 2√3) = 0
x = 3i, 2√3
iv) 6x 2 – 17ix + 12i 2 = 0
Solución:
6x 2 – 9ix – 8ix + 12i 2 = 0
3x(2x – 3i) – 4i(2x – 3i) = 0
x = 4/3i, 3/2i
Pregunta 2. Resuelve las siguientes ecuaciones cuadráticas:
i) x 2 – (3√2 + 2i)x + 6√2i = 0
Solución:
x 2 – 3√2x – 2ix + √2i = 0
x(x – 2i)(x – 3√2) = 0
(x – 2i)(x – 3√2) = 0
x = 2i, 3√2
ii) x 2 – (5 – i)x + (18 + i) = 0
Solución:
x 2 – 5x – ix +18 + i = 0
x2 – (3 – 4i)x – (2 + 3i)x + (18 + i) = 0
x(x – (3 – 4i)) – (2 + 3i)x + (18 + i) = 0
(x – (2 + 3i)) (x – (3 – 4i)) = 0
x = 2 + 3i, 3 – 4i
iii) (2 + i)x 2 – (5 – i)x + 2(1 – i) = 0
Solución:
(2 + i)x 2 – 2x – (3 – i)x + 2(1 – i) = 0
x[(2 + i)x – 2] – (1 – i)[(2 + i)x – 2] = 0 (tenemos, i 2 = -1)
[x – (1 – i)] = 0 o [(2 + i)x- 2] = 0
x = 1 – yo o x = 2 / 2 + yo
más podemos extender x = 2 / 2 + i
x = 2 / 2 + yo * 2 – yo / 2 – yo
x = 4 – 2i / 4 + 1
x = 4 / 5 – 2 / 5i
iv) x 2 – (2 + i)x – (1 – 7i) = 0
Solución:
x2 – ( 2 + i)x – (1 – 7i) = 0
x 2 – (3 – i)x + (1 – 2i)x – (1 – 7i) = 0
x(x – (3 – i)) + (1 – 2i)(x – (3 – i)) = 0
[x + (1 – 2i)] [x – (3 – i)] = 0
x = -1 + 2i, 3 – yo
v) ix 2 – 4x – 4i
Solución:
ix 2 + 4i 2 + 4i 3 = 0 [ i 2 = -1; (4i 3 ) = 4i(i 2 ) = 4i(-1) = -4i ]
i(x^2 + 2ix + 2ix + 4i 2 ) = 0
x^2 + 2ix + 2ix + 4i 2 = 0
x(x + 2i) + 2i(x + 2i) = 0
x = -2i, -2i
vi) x 2 + 4ix – 4 = 0
Solución:
x2 + 4ix + 4i2 = 0 [ i2 = -1]
x 2 + 2ix + 2ix + 4i 2 = 0
x(x + 2i) + 2i(x + 2i) = 0
x = -2i, -2i
vii) 2x 2 + √15ix – i = 0
Solución:
Aquí usamos la forma general de la ecuación cuadrática
Comparando con la forma general de la ecuación cuadrática ax 2 + bx + c
a = 2, b = √15i, c = -i; sustituir estos valores en la fórmula
r1 = (-b + √b 2 -4ac)/2a; r2 = (-b – √b 2 -4ac)/2a
r1 = (- √15i + √-15 + 8i) / 4; r2 = (-√15i – √-15 + 8i) / 4
sea √-15 + 8i = a + bi
=> -15 + 8i = (a + bi) 2
=> -15 + 8i = a 2 – b 2 + 2abi
=> a 2 – b 2 = – 15 y 2abi = 8i
tenemos, (a 2 + b 2 ) 2 = ((a 2 – b 2 ) 2 ) + 4 * a 2 * b 2
(a 2 + b 2 ) 2 = (-15) 2 + 64 = 289
un 2 + segundo 2 = 17
Resolviendo a 2 – b 2 = -15 y a 2 + b 2 = 17 obtenemos
a 2 = 1 y b 2 = 16
a = 1,b = 4 o a = -1,b = -4
entonces, √-15 + 8i = 1 + 4i, -1 – 4i
cuando √-15 + 8i = 1 + 4i
r1 = (-√15i + 1 + 4i) / 4 = 1 + (4 – √15) i / 4
r2 = ( -√15i -(1 + 4i) ) / 4 = -1 – (4 + √15)i / 4
cuando √-15 + 8i = -1 – 4i
r1 = (-√15i – 1 – 4i) / 4 = -1 – (4 + √15)i / 4
r2 = (-√15i – (-1 – 4i) ) / 4 = 1 + (4 – √15)i / 4
viii) x 2 – x + (1 + i) = 0
Solución:
x 2 – x + (1 + yo) = 0
x 2 – ix – (1 – i)x + i(1 – i) = 0
(x – i)(x – (1 – i)) = 0
x = yo,1 – yo
ix) ix 2 – x + 12i = 0
Solución:
Aplicación de la regla de discriminación
x = (-b ±√b 2 – 4ac) / 2a
x = -(-1) ± ( √(-1) 2 – 4i * 12i) / 2i
=1 ± √1 + 48/ 2i
= 1 ± √49 / 2i
=1 ± 7 / 2i
= 8 / 2i, -6 / 2i
= 4 / yo, -3 / yo
= -4i, 3i
x) x2 – (3√2 – 2i)x – √2i = 0
Solución:
Aplicación de la regla de discriminación
x = (-b + √b 2 – 4ac) / 2a
x = ((3√2 – 2i) ± √[-(3√2 – 2i)] 2 – 4 (-√2i)) / 2
= ((3√2 – 2i) ± √(3√2 – 2i) 2 + 4√2i) / 2
= (3√2 – 2i) / 2 ± ( 4√2i) / 2
xi) x2 – (√2 + i)x + √2i = 0
Solución:
x 2 – √2x – ix + √2i = 0
x(x – √2) – i(x – √2) = 0
(x – i)(x – √2) = 0
x = yo, √2
xii) 2x 2 – (3 + 7i)x + (9i – 3) = 0
Solución:
2x 2 – 3x – 7ix + (9i – 3) = 0
(2x – 3 – i)(x – 3i) = 0
(x – 3 + i / 2) (x – 3i) = 0
x = 3 + i/2, 3i
Publicación traducida automáticamente
Artículo escrito por vishnuteja476 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA