first line: 35-x^2+2x=0
next important line: x^2-2x-35 = 0 , I don't know why you did all that other stuff in between.
(see my response to a similar problem in the reply to "Lori" two postings down from here)
so x^2-2x-35 = 0
(x-7)(x+5) = 0
then x=7 or x = -5
you had as factors (x-7)(x+5) (x-7)(x+5).
Why 4 of them?? It was only a quadratic, so it could have at most two factors.
that was from the example we were given, I just followed it. Our instructor was ill and the sub handed out examples, no one understood and the sub could not explain! Sorry I have the correct answers, but I need to go back and show my work the right way? I will look in my book. Thanks
Here is all you would have needed, down to the essential steps
35-x^2+2x=0 (multiply by -1 to get it in standard form
x^2 - 2x - 35 = 0 (this factors to)
(x-7)(x+5) = 0
therefore x-7 = 0 or x+5 = 0
therefore x = 7 or x = -5
If you are asked to verfy the solution, this is how you would do that:
if x=7, then
L.S. = 7^2 - 2(7) - 35
= 49 - 14 - 35 = 0
if x= -5
L.S. = (-5)^2 - 2(-5) - 35
= 25 + 10 - 35
Therefore x = 7 or x = -5
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