Can you Solve Cambridge University Admission Test ?✍️🖋️📘💙

[@2012tulio]

x= 2 sqrt i x= -2 sqrt i x = 2 sqrt i^3 x = -2 sqrt i^3

[@payoo_2674]

x⁴=-16 Let x=a+bi {a,b∈R} (a+bi)⁴=-16 a⁴+4a³bi-6a²b²-4ab³i+b⁴=-16+0i Comparing the real and imaginary parts of both sides of the equation a⁴-6a²b²+b⁴=-16 ① and 4a³b-4ab³=0 4ab(a²-b²)=0 4ab(a-b)(a+b)=0 => 4 cases case 1: a=0 (rejected: from ① b⁴=-16, but b∈R) case 2: b=0 (rejected: from ① a⁴=-16, but a∈R) case 3: a-b=0 => a=b => from ①: -4b⁴=-16 => b⁴=4 => b=±⁴√4=±√2 (complex roots rejected, b/c b∈R) if b=√2 => a=√2 => x₁=√2+i√2 if b=-√2 => a=-√2 => x₂=-√2-i√2 case 4: a+b=0 => a=-b => from ①: -4b⁴=-16 => b⁴=4 => b=±⁴√4=±√2 (complex roots rejected, b/c b∈R) if b=√2 => a=-√2 => x₃=-√2+i√2 if b=-√2 => a=√2 => x₄=√2-i√2

[@payoo_2674]

x⁴=-16 x⁴+16=0 (x²)²-(4i)²=0 (x²-4i)(x²+4i)=0 (x²-(2√i)²)(x²-(2i√i)²)=0 but √i=1/√2+i/√2 (principal root) (x-2(1/√2+i/√2))(x+2(1/√2+i/√2))(x-2i(1/√2+i/√2))(x+2i(1/√2+i/√2))=0 (x-√2-i√2)(x+√2+i√2)(x+√2-i√2)(x-√2+i√2)=0 so x₁=√2+i√2 x₂=-√2-i√2 x₃=-√2+i√2 x₄=√2-i√2

[@subratabiswas2502]

x^4=16i^2 ; i,e x^4-16i^2=0 ; i,e (x^2+4i)(x^2-4i) ; So, i= (1/2)×2i = (1/2)[1+2i-i^2]= (1+i)^2 ; so, root(i)=(1/root2)(1+i); so, root(4i)=+2×1/root2(1+i)=+root2(1+i) and -root2(1+i) ; similarly, root(-4i)=+root2(1-i) and -root2(1-i). x=root(4i) & x= root(-4i). Hence, x= +root2(1+i), -root2(1-i). , +root2(1-i) or -root2(1-i).