Question

If ${{\left( x-1 \right)}^{4}}-16=0$, then the sum of non- real complex value of x is:
(a). 2
(b). 0
(c). 4
(d). None of these

Answer 1

Hint: We have to find the complex root of the given equation. The formula that we will use here is ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ , after finding the two complex root we will add them and that will be the final answer.

Complete step-by-step answer:

Let’s start our solution.
We have been given ${{\left( x-1 \right)}^{4}}-16=0$
Taking 16 on the other side we get,
${{\left( x-1 \right)}^{4}}=16$
Now we need to find the complex root,
Now we can write 16 as ${{\left( \pm 2i \right)}^{4}}$ because ${{\left( \pm 2 \right)}^{4}}=16$ and we also know that ${{i}^{4}}=1$
Hence, using this two relation we get,
$\begin{align}
  & {{\left( x-1 \right)}^{4}}={{\left( \pm 2i \right)}^{4}} \\
 & x-1=\pm 2i \\
 & x=1\pm 2i \\
\end{align}$
Hence, we have found the two complex root of the equation as 1 + 2i and 1 – 2i
Now adding the complex root we get,
$1+2i+1-2i=2$
Hence, the sum of the complex root of the equation ${{\left( x-1 \right)}^{4}}-16=0$ is 2.
Hence, the correct option is (a).

Note: We can see that the coefficients of the equation are real so the complex root will always occur in a conjugate pair. Conjugate pair of a + ib = a – ib, now the highest power of x is 4, hence the total number of roots will also be 4. We can also see that 2 and -2 are the real roots and hence the other two roots will be complex and it will be a conjugate pair.