Question 107338


An imaginary number is represented as {{{ai}}} where {{{a}}} is any positive or negative number, and {{{i}}} is a {{{special }}} {{{imaginry}}}{{{constant }}}whose square equals {{{-1}}} (that means {{{i*i = -1}}}).  

So, we may then take the square root out of any negative number 
{{{sqrt(-x) = sqrt(x)*sqrt(-1) = sqrt(x)*i}}}.

 

{{{3sqrt(-5)(sqrt(-3)+ 4sqrt(-5))}}}

{{{3sqrt(5)sqrt(-1)(sqrt(3)sqrt(-1)+ 4sqrt(5)sqrt(-1))}}}

{{{3sqrt(5)*i(sqrt(3)*i+ 4sqrt(5)*i)}}}

{{{3sqrt(5)*i*i(sqrt(3)+ 4sqrt(5))}}}

{{{3sqrt(5)*(-1)(sqrt(3)+ 4sqrt(5))}}}

{{{-3sqrt(5)(sqrt(3)+ 4sqrt(5))}}}


{{{(-(3sqrt(5))(sqrt(3)))+ (-3sqrt(5)4sqrt(5))}}}

{{{(-(3sqrt(5))(sqrt(3)))+ (-12sqrt(5)sqrt(5))}}}

{{{(-(3sqrt(5))(sqrt(3)))+ (-12sqrt(5)^2)}}}

{{{(-(3sqrt(5))(sqrt(3)))+ (-12*5)}}}




now, you can plug in values for {{{sqrt(5)}}} and {{{sqrt(3)}}} to find final result