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algebra

posted by on .

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 - ,

    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 - ,

    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 - ,

    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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