SOLUTION: I'm having a lot of trouble with a question on one of my Algebra two assignments and was wonder if could get some help with it. 10. -57 3/5 x2 + 3 3/5 x = -39 3/5 x2

Click here to see ALL problems on Complex Numbers

Question 323969: I'm having a lot of trouble with a question on one of my Algebra two assignments and was wonder if could get some help with it.
10. -57 3/5 x2 + 3 3/5 x = -39 3/5 x2 - 7 1/5 x + 14 2/5
A x = {2 +- i[SQRT(75)}/10
B x = {3 +- i[SQRT(71)]}/-10
C x = {-3 +- i[SQRT(57)]}/7
D x = {-3 +- i[SQRT(71)]}/-10
E x = {-2 +- i[SQRT(75)]}/-7
F x = {2 +- i[SQRT(39)]}/14

Any help at all would be greatly appreciated
Thanks you
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
-57 3/5 x2 + 3 3/5 x = -39 3/5 x2 - 7 1/5 x + 14 2/5
Multiply by 5

Divide by 18

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -71 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -71 is + or - .

The solution is , or
Here's your graph:

------------------------
Looks like it's D, but the 2 minus signs can be eliminated.

SOLUTION: I'm having a lot of trouble with a question on one of my Algebra two assignments and was wonder if could get some help with it. 10. -57 3/5 x2 + 3 3/5 x = -39 3/5 x2 - 7 1/5 x +