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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Quadratic Equations and Parabolas
-> 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):
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....
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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