Results 1 to 5 of 5

Math Help - Completely stumped on this problem!

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    2

    Completely stumped on this problem!

    If x squared = 25, y squared = 16, and z squared = 9 what is the greatest possible value of (x + y - z)squared?

    My book says the answer is 144. Can someone please tell me how that the book got this answer?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by freddy View Post
    If x squared = 25, y squared = 16, and z squared = 9 what is the greatest possible value of (x + y - z)squared?

    My book says the answer is 144. Can someone please tell me how that the book got this answer?
    x = \pm  5 , y = \pm 4 , and z = \pm 3

    you want x , y , and -z to all have the same sign to maximize the square

    [5 + 4 - (-3)]^2 = 12^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    2

    how?

    How can you make the 3 negative if its positive in the problem? You have 2 minus signs yet there is only 1 minus sign in the problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by freddy View Post
    How can you make the 3 negative if its positive in the problem? You have 2 minus signs yet there is only 1 minus sign in the problem.
    freddy ...

    -(-3) = +3
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie gs.sh11's Avatar
    Joined
    Oct 2009
    Posts
    14
    When you take the square root of any number such as 25 a negative can also be the solution, for example (-5)(-5) = 25, as (5)(5) = 25, so x = -5, y = -4, and because the equation is (x + y - z), z = 3, so now

    (-5 - 4 - 3)^2 = 144
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with SPSS stats...completely stumped
    Posted in the Statistics Forum
    Replies: 0
    Last Post: October 21st 2011, 08:58 AM
  2. Replies: 13
    Last Post: April 10th 2010, 06:15 AM
  3. Completely stumped by a picture frame >_<
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 31st 2009, 07:48 AM
  4. Replies: 11
    Last Post: June 30th 2008, 07:26 PM
  5. Im completely stumped by this one.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 23rd 2008, 03:44 PM

Search Tags


/mathhelpforum @mathhelpforum