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
[tex]\sqrt{D} = \sqrt{-12}=\sqrt{2*2*3i^{2}}=2i\sqrt{3}[/tex]
root = (-b ± √D)/2a
= (-(-2) ± 2i√3)/2*1
= (2 ± 2i√2)/2
[tex]= \dfrac{2(1 + i\sqrt{3})}{2} \ ; \dfrac{2(1+i\sqrt{3})}{2}\\\\\\= 1 + i\sqrt{3} \ ; \ 1 - i\sqrt{3}[/tex]
