SOLUTION: 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

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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
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Try it this way:



and then INSIDE the parentheses, add the term , but on the right-hand member, you must account for the factor in the left-side member which has been factorized....




If that is clear for you, then continue.
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
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
 

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