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Posted by on Friday, January 13, 2012 at 6:29pm.

how can you solve x^2 - 3x = 0 i was using a formula as:

x^2-7x+10=0
(x-5)(x+2)= 0
x-5=0
x+5 +5 x=5

I was following this formula but got stumped on this question please help.

  • integrated algebra - , Friday, January 13, 2012 at 6:31pm

    x (x-3) = 0
    x = 0 or x = 3

  • integrated algebra - , Friday, January 13, 2012 at 6:57pm

    before looking at your answer i found the answer to be 0 too but like this:

    x^2-3x=0
    x^2-3x+0=0
    (x-0)(x+0)=0
    0^2-3(0)=0

    was my way right

  • integrated algebra - , Friday, January 13, 2012 at 7:10pm

    No, because -0x-0x is not -3x
    You missed the x = 3 root entirely.
    Please do it my way :)

  • integrated algebra - , Friday, January 13, 2012 at 7:15pm

    If you insist on completing the square:

    x^2 - 3x = 0

    (-3/2)^2 = 9/4
    so
    x^2 - 3 x + 9/4 = 9/4

    (x-3/2)^2 = 9/4 =3/2

    x- 3/2 = +/- 3/2

    x = 3/2 + 3/2 = 3
    pr
    x = 3/2 - 3/2 = 0

  • integrated algebra - , Friday, January 13, 2012 at 7:18pm

    by the way

    x^2-7x+10=0
    (x-5)(x+2)= 0
    x-5=0
    x = 5
    OR
    x+2 = 0
    x = -2

  • integrated algebra - , Friday, January 13, 2012 at 7:25pm

    do i always have to mention the ors

  • integrated algebra - , Friday, January 13, 2012 at 7:36pm

    If it is a quadratic, in general it has two solutions, so you need to give both.
    Sometimes, if the vertex of the parabola is exactly on the x axis, the two roots will be the same. For example:
    x^2 -4x + 4 = 0
    (x-2)(x-2) = 0
    x = 2
    or
    x = 2

  • integrated algebra - , Friday, January 13, 2012 at 7:54pm

    oh, okay, thank you!

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