Question 1059706
Solve the equation by completing the square and applying the square root property. Write imaginary solutions in the form a+bį.
2a^2+4a+5=0 
2a^2+4a=-5 
((2a^2)/2)+(4a/2))=(-5/2) 
(1/2)(2)^2) = 1
a^2+2a+1=-5/2+1
(a+1)^2=(-5/2)+(2/2)
(a+1)^2=-3/2
sqrt (a+1)^2=sqrt(-3/2)
a+1=+or- sqrt(3/2)į
<pre>{{{2a^2 + 4a + 5 = 0}}} 
{{{2a^2 + 4a = - 5}}} <======== Correct
{{{2a^2/2 + (4a/2) = (- 5)/2}}} <======= Correct
{{{((1/2) * 2)^2 = 1}}} <======= Correct
{{{a^2 + 2a + 1 = (- 5)/2 + 1}}} <======= Correct
I must say, so far, I'm impressed
{{{(a + 1)^2 = (- 5)/2 + 2/2}}} <===== Correct
{{{(a + 1)^2 = (- 3)/2}}} <====== Correct
{{{sqrt((a + 1)^2) = " "+-sqrt(- 3/2)}}}
{{{a + 1 = " "+-sqrt(- 3/2)}}} 
{{{a + 1 = " "+- sqrt(3/2)i}}} <====== Correct
{{{a + 1 = " "+- (sqrt(3) * sqrt(2)/2) * i}}} =====> {{{a + 1 = " "+- (sqrt(6)/2)i}}} =====> {{{highlight(highlight_green(highlight(a = - 1 +- (sqrt(6)/2)i)))}}}