Posted by Lisa on Friday, March 21, 2008 at 6:30pm.
This is how my book shows it, but I don't know
solve the equation by factoring
35x^2+2x=0
x^2 x^2
35+2x=x^2
2x=2x
3535=x^22x35
0=x^22x35
(x7)(x+5) (x7)(x+5)
x7=0 x+5=0
x=7 x=5
(7)^22(7)35 5^22(5)35
491435 25+1035
0 0
Solution set is {5,7}

algebra  Reiny, Friday, March 21, 2008 at 6:42pm
first line: 35x^2+2x=0
next important line: x^22x35 = 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^22x35 = 0
(x7)(x+5) = 0
then x=7 or x = 5
you had as factors (x7)(x+5) (x7)(x+5).
Why 4 of them?? It was only a quadratic, so it could have at most two factors.

algebra  Anonymous, Friday, March 21, 2008 at 6:48pm
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

algebra  Reiny, Friday, March 21, 2008 at 6:56pm
Here is all you would have needed, down to the essential steps
35x^2+2x=0 (multiply by 1 to get it in standard form
x^2  2x  35 = 0 (this factors to)
(x7)(x+5) = 0
therefore x7 = 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
= R.S.
if x= 5
L.S. = (5)^2  2(5)  35
= 25 + 10  35
= 0
= R.S.
Therefore x = 7 or x = 5
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