Question 172054
The additive inverse of any number is that number when added to the original number results in a sum of zero.  So if {{{a}}} is your original number, the additive inverse is {{{-a}}} because {{{a+(-a)=0}}} for all {{{a}}}.


To sum two complex numbers, add the real parts and add the imaginary parts, thus:


{{{(a+bi)+(c+di)= (a+c)+(b+d)i}}}


So, in order to find the additive inverse of {{{-4+6i}}} we need to determine the additive inverses of both {{{-4}}} and {{{6}}}, namely {{{4}}} and {{{-6}}}.  Hence the additive inverse of {{{-4+6i}}} is {{{4-6i}}} because


{{{(-4+6i)+(4-6i)=0+0i=0 }}}


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The absolute value of a complex number {{{z=(x+yi)}}} is given by {{{abs(z)=sqrt(x^2+y^2)}}}.


In this case, you are looking for {{{abs(6i)=abs(0+6i)=sqrt(0^2+6^2)=sqrt(36)=6}}}


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{{{x = (-b +- sqrt( b^2-4ac ))/(2a) }}}  are the two roots of a quadratic equation of the form {{{ax^2+bx+c=0}}}


All you need to do is plug in the values of the coefficients in your quadratic equation and do the arithmetic.  Remember to use a factor of i if the part under the radical is less than zero.


For your first equation, {{{a=-1}}}, {{{b=5}}}, and {{{c=-2}}}, so:


{{{x = (-5 +- sqrt( 5^2-4(-1)(-2) ))/(2(-1)) }}}



{{{x = (-5 +- sqrt( 25-8 ))/(-2) }}}



{{{x = (-5 +- sqrt( 17 ))/(-2) }}}


Hence {{{x[1]=(5-sqrt(17))/2}}} and {{{x[2]=(5+sqrt(17))/2}}}


These are the exact representations of the two roots of this quadratic.  If you want numerical approximations, use your calculator.  Seek guidance from your text or your instructor as to the required precision whenever numerical approximations are required.


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The other problem is worked exactly the same way, and again has a positive discriminant, the second one also has real number roots.