Question

if $(1-x{)}^4-16=0$ then the sum of nonreal complex values of x is

• A. $2$
• B. $0$
• C. $4$
• D. None of these

Answer 1

Answer: (A)

$2$

Given, $(1-x{)}^4-16=0$
$\Rightarrow (x-1{)}^4=16=4^2$
$\Rightarrow (x-1{)}^2=\pm 4=2^2,(2i{)}^2$, where $i^2=-1$
$\Rightarrow (x-1)=\pm 2,\pm 2i$
$\Rightarrow x=1\pm 2,1\pm 2i$
Hence sum of non real complex value of $x$ is $=(1+2i)+(1-2i)=2$