SOLUTION: Find all complex-number solutions
2x^2-3x-5=0
I just need to see steps on how to break this down. I know the first part is adding 5 to make C.
Question 1027646: Find all complex-number solutions
2x^2-3x-5=0
I just need to see steps on how to break this down. I know the first part is adding 5 to make C.
If you EXPECT complex solutions, then either solve using Completing the Square, or go directly to the general solution of a quadratic formula. You are maybe intent on Completing The Square, since you mention "first part is adding 5...".
You should recognize what that was....
...but, multiply left and right members by ... -------solution will not contain an Imaginary part.
As a way of checking quality of the result, find the discriminant on the original equation. , A POSITIVE NUMBER, so you expect two REAL solutions for x. Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
Find all complex-number solutions
2x^2-3x-5=0
I just need to see steps on how to break this down. I know the first part is adding 5 to make C.
Why would you want to add 5. Are you supposed to solve this by completing the square? If not, the trinomial can be factored as:
(2x - 5)(x + 1) = 0
2x - 5 = 0 OR x + 1 = 0 . You are done!!
As seen there are NO COMPLEX-NUMBER solutions.