Question 1207475
<font color=black size=3>
Answer: <font color=red>4+i</font>


Quick explanation:  If a+bi is one root, then its paired counterpart (known as the complex conjugate) would be a-bi
This applies only when all of the coefficients are real numbers.


Another example: The root 2+i pairs up with 2-i


-----------------------------------------------------------------------------------------------------------------------------
-----------------------------------------------------------------------------------------------------------------------------


Let's say you didn't know about the complex conjugate, or that you might be curious about an alternative pathway.



We can isolate the "i" term and square both sides to generate a quadratic from it.
{{{x = 4-i}}}


{{{x-4 = -i}}}


{{{(x-4)^2 = (-i)^2}}}


{{{(x-4)^2 = -1}}}


{{{x^2-8x+16 = -1}}}


{{{x^2-8x+16+1 = 0}}}


{{{x^2-8x+17 = 0}}}


Now let's apply the quadratic formula
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-8)+-sqrt((-8)^2-4(1)(17)))/(2(1))}}}


{{{x = (8+-sqrt(64 - 68))/(2)}}}


{{{x = (8+-sqrt(-4))/(2)}}}


{{{x = (8+- 2i)/(2)}}}


{{{x = (8+ 2i)/(2)}}} or {{{x = (8-2i)/(2)}}}


{{{x = 4+i}}} or {{{x = 4-i}}}
We arrive at roots <font color=red>4+i</font> and 4-i to help confirm the answer. 


This is a somewhat long-winded pathway to basically rephrase what I mentioned at the top. 
If a+bi is one root, then a-bi is also included in the mix. 
This only applies when all coefficients are real numbers. 
</font>