Question

Q.if X=√5 iota -2

then,, what is value of X^4+X^3+X^2+X+5?

Answer 1

Solution to your question is as
under :-

Given value of x = √5i-2
Given equation = x⁴+x³+x²+x+5

First of all we will find the values of all the terms given in the equation and then we will substitute these values in the equation itself.

x² = ( √5i-2 )²
= ( 5i²+ 4 - 4√5i )
= ( -5 + 4 - 4√5i ) { Since, i = √-1,
Therefore i² = -1 }
= ( -1 - 4√5i )


x³= x².x
= ( -1 - 4√5i )(√5i - 2 )
= ( -√5i + 2 + 20 + 8√5i )
= ( 22 + 7√5i )


x⁴= x². x²
= ( -1 - 4√5i )( ( -1 - 4√5i )
= 1 + 4√5i + 4√5i + 80i²
= 1 + 8√5i - 80
= -79 + 8√5i

Now, we have all the values, just substitute them in the given equation.

Equation is :- x⁴+x³+x²+x+5
= -79 + 8√5i +22 + 7√5i - 1 - 4√5i + √5 i -2 + 5
= -79 + 22 -1 + 3 + 8√5i + 7√5i - 4√5i + √5i
= - 55 + 12√5i

So, the answer to your question is

= - 55 + 12√5i


Hope it helps you !
Thank You !