Question 139800

{{{z^2-2z+1=-4}}} Start with the given equation



{{{z^2-2z+1+4=0}}}  Add 4 to both sides. 



{{{z^2-2z+5=0}}} Combine like terms



Let's use the quadratic formula to solve for z:



Starting with the general quadratic


{{{az^2+bz+c=0}}}


the general solution using the quadratic equation is:


{{{z = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{z^2-2*z+5=0}}} ( notice {{{a=1}}}, {{{b=-2}}}, and {{{c=5}}})





{{{z = (--2 +- sqrt( (-2)^2-4*1*5 ))/(2*1)}}} Plug in a=1, b=-2, and c=5




{{{z = (2 +- sqrt( (-2)^2-4*1*5 ))/(2*1)}}} Negate -2 to get 2




{{{z = (2 +- sqrt( 4-4*1*5 ))/(2*1)}}} Square -2 to get 4  (note: remember when you square -2, you must square the negative as well. This is because {{{(-2)^2=-2*-2=4}}}.)




{{{z = (2 +- sqrt( 4+-20 ))/(2*1)}}} Multiply {{{-4*5*1}}} to get {{{-20}}}




{{{z = (2 +- sqrt( -16 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{z = (2 +- 4*i)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{z = (2 +- 4*i)/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{z=1 + 2*i}}} or {{{z=1 - 2*i}}}