Question

Verify that is a root of the equation and determine the other roots of the equation

Answer 1

Jawab:

Penjelasan dengan langkah-langkah:

(-2+3i)³+5(-2+3i)²+17(-2+3i)+13

= (-2+3i)²(-2+3i+5)-34+51i+13

= (4-9-2.2.3i)(3+3i)-21+51i

= -3(5+12i)(1+i)-21+51i

= -3(-7+17i)-21+51i

= 21-51i-21+51i

= 0 (proven)

z³+5z²+17z+13 = (z+(2-3i))(z²+az+b)

b.(2-3i) = 13, b = 13/(2-3i) = 13(2+3i)/(13) = 2+3i

z³+5z²+17z+13 = (z+(2-3i))(z²+az+(2+3i))

                       = z³+(a+2-3i)z²+(2+3i+a(2-3i))z+13

z³+5z²+17z+13 = z³+(a+2-3i)z²+(2(a+1)+3i(1-a))+13

compare coefficient :

a+2-3i = 5

a = 3(1+i)

2(a+1)+3i(1-a) = 17

2a+2+3i-3ia = 17

a(2-3i) = 15-3i

a = (15-3i)/(2-3i)

a = (15-3i)(2+3i)/13

a = (30+9+45i-6i)/13

a = (39+39i)/13

a = 3(1+i) (the same √ )

z³+5z²+17z+13 = (z+(2-3i))(z²+3(1+i)z+(2+3i))

to determine 2 other roots use the quadratic :

z = (-3(1+i) ± √(9(1+i)²-4.1.(2+3i)) )/(2.1)

  = (-(3+i) ± √(9.2i-8-12i))/(2)

z  = (-(3+i) ± √(6i-8))/(2)

z = -(3+i)/(2) ± √(3i/2 - 2)

z = -(3+i)/(2) + √(3i/2 - 2) or -(3+i)/(2) - √(3i/2 - 2)