Results 1 to 6 of 6

Math Help - Need help with an equation

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    7

    Need help with an equation

    Can anyone finish this equation?

    (10)^2=(x/2)^2+((x+4)/2)^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2008
    Posts
    7
    I know that the next step looks like this, but I don't understand where the 2x that is underlined came from??

    Would appreciate it if anyone could answer that question


    100=x^2/4+x^2/4+2x+4
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by Yaymath View Post
    Can anyone finish this equation?

    (10)^2=(x/2)^2+((x+4)/2)^2
    Not so bad...

    10^2=\left(\frac{x}{2}\right)^2+\left(\frac{x+4}{2  }\right)^2

    100=\frac{x^2}{4}+\frac{x^2+8x+16}{4}

    400=x^2+x^2+8x+16

    2x^2+8x-384=0

    x^2+4x-192=0

    (x-12)(x+16)=0

    x=12 \ \ or \ \ x=-16
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2008
    Posts
    7
    I understand everything except the last part







    (did you do the additive inverse of 400 or 16?)

    (how did this become to (x - 12)(x+16)=0





    The rest is clear for me, it's just there i'm confused
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by masters View Post
    Not so bad...

    10^2=\left(\frac{x}{2}\right)^2+\left(\frac{x+4}{2  }\right)^2

    100=\frac{x^2}{4}+\frac{x^2+8x+16}{4}

    400=x^2+x^2+8x+16

    2x^2+8x-384=0

    x^2+4x-192=0

    I factored the above equation. You could have used the quadratic formula if you wanted to get the values for x.

    (x-12)(x+16)=0

    I used the zero product property here. The zero product property says that if ab=0, then a=0 or b=0.

    Either x - 12= 0 or x + 16 = 0

    Then I solved each equation to get:

    x=12 \ \ or \ \ x=-16
    ..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2008
    Posts
    7
    Thanx alot you are indeed the master

    Bye and thnx again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 11th 2011, 01:17 AM
  2. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 11:39 AM
  3. Replies: 2
    Last Post: May 18th 2009, 12:51 PM
  4. Replies: 2
    Last Post: April 28th 2009, 06:42 AM
  5. Replies: 1
    Last Post: October 23rd 2008, 03:39 AM

Search Tags


/mathhelpforum @mathhelpforum