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

35-x^2+2x=0

x^2 x^2

35+2x=x^2

-2x=-2x

35-35=x^2-2x-35

0=x^2-2x-35

(x-7)(x+5) (x-7)(x+5)

x-7=0 x+5=0

x=7 x=-5

(7)^2-2(7)-35 -5^2-2(-5)-35

49-14-35 25+10-35

0 0

Solution set is {-5,7}

- algebra -
**Reiny**, Friday, March 21, 2008 at 6:42pm
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.

- 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

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

= 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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