Question

PLEASE HELP WITH THIS ONE QUESTION

Answer 1

Answer:

x = 0 , -2 , 1 +i√3  ; 1 - i√3

Step-by-step explanation:

2x⁴ + 16x = 0

2x(x³ + 8) = 0

2x(x³ + 2³)  = 0

2x (x + 2)(x² -2x + 2²) = 0    {a³ + b³ = (a + b)*(a² -ab +b²}

2x = 0  

x = 0

x + 2 = 0

x = -2

x² - 2x + 4 = 0

a = 1 ; b = -2 and c = 4

D = b² - 4ac = (-2)² - 4*1*4= 4 - 16 = -12

$\sqrt{D} = \sqrt{-12}=\sqrt{2*2*3i^{2}}=2i\sqrt{3}$

root = (-b ± √D)/2a

= (-(-2) ± 2i√3)/2*1

= (2 ± 2i√2)/2

$= \dfrac{2(1 + i\sqrt{3})}{2} \ ; \dfrac{2(1+i\sqrt{3})}{2}\\\\\\= 1 + i\sqrt{3} \ ; \ 1 - i\sqrt{3}$