Results 1 to 11 of 11

Math Help - Problem Solving

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    134

    Problem Solving

    I have 6 problems here that I cannot solve... Help me please. These are problems that require factoring... I need the solutions also... Here are the problems..

    1.) Find two integers which differ by 5, while their squares differ by 85.

    4.) A rectangular swimming pool with dimensions of 11m. by 8m. is built in
    rectangular backyard 460m2. If the strip of yard surrounding the is of
    uniform width, how wide is the strip?

    5.) If the sides of a square are lengthened by 5cm. the area becomes 64cm2.
    Find the length of the side of the original square.

    6.) The pages of a book are 15cm. by 20cm. Margin of equal width surround the
    print on each page and comprise one half the area of the page. Find the
    width of the margins.

    7.) The sides of a cube is 2cm. more than the side of another cube. The volumes of
    the cubes differ by 152cm2. Find the length of the sides of each cube.

    8.) The product of two consecutive positive integers is 210. Find the integers.

    THANK YOU IN ADVANCE!
    Last edited by reiward; September 10th 2006 at 06:27 PM. Reason: forgot something
    Follow Math Help Forum on Facebook and Google+

  2. #2
    dan
    dan is offline
    Member dan's Avatar
    Joined
    Jun 2006
    From
    nebraska usa
    Posts
    113
    Quote Originally Posted by reiward View Post
    8.) The product of two consecutive positive integers is 210. Find the integers.
    [tex] x(x+1)=210[tex]
    [tex] x^2 + x - 210=0 [tex]

    use the quadractic formula...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    134
    x^2+x-210 = 0
    that's...
    (x-14) | (x+15)
    x-14=0 | x+15=0
    x=14 | x=-15

    They're not both positives?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    dan
    dan is offline
    Member dan's Avatar
    Joined
    Jun 2006
    From
    nebraska usa
    Posts
    113
    Quote Originally Posted by reiward View Post
    5.) If the sides of a square are lengthened by 5cm. the area becomes 64cm2.
    Find the length of the side of the original square.
    if the original side is x

    then (x+5)^2 = 64
    or x^2 + 10x + 25=64
    or x^2 + 10x -39 =0

    use the quadradic fromula again...

    dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2006
    Posts
    134
    x^2+10x-39=0
    (x-3) (x+13)
    x=3 | x=-13

    Okay help please.. I always end up having positive and negative answers....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    dan
    dan is offline
    Member dan's Avatar
    Joined
    Jun 2006
    From
    nebraska usa
    Posts
    113
    Quote Originally Posted by reiward View Post
    x^2+x-210 = 0
    that's...
    (x-14) | (x+15)
    x-14=0 | x+15=0
    x=14 | x=-15

    They're not both positives?

    by the quadradic formula x=14,x=15 then x+1 =15 or (x-1)=14


    ether way your values are 14 and 15 and 14*15=210
    dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by reiward View Post
    x^2+10x-39=0
    (x-3) (x+13)
    x=3 | x=-13

    Okay help please.. I always end up having positive and negative answers....
    And?
    You reject the negative. Thus there is only one answer.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    dan
    dan is offline
    Member dan's Avatar
    Joined
    Jun 2006
    From
    nebraska usa
    Posts
    113
    Quote Originally Posted by reiward View Post
    x^2+10x-39=0
    (x-3) (x+13)
    x=3 | x=-13

    Okay help please.. I always end up having positive and negative answers....
    because you are using the quadradic formula you have the "+/- square root" so you will usually have two values...usally only one fits the problem though.
    dan
    Last edited by ThePerfectHacker; September 10th 2006 at 06:50 PM. Reason: dan made mistake
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2006
    Posts
    134
    alright, i get it.. next problms please
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by reiward View Post

    1.) Find two integers which differ by 5, while their squares differ by 85.
    You have,
    x-y=5
    x^2-y^2=85

    Factor the second,
    x-y=5
    (x-y)(x+y)=85

    Substitute,
    x-y=5
    5(x+y)=85

    Thus,
    x-y=5
    x+y=17

    Add equations,
    2x=22

    Thus,
    x=11

    Thus,
    y=6
    Quote Originally Posted by reiward
    5.) If the sides of a square are lengthened by 5cm. the area becomes 64cm2.
    Find the length of the side of the original square.
    The sides of the new square must be 8 cuz 8^2=64.
    Thus, the old square must have had 3 cuz 3+5=8.
    Last edited by ThePerfectHacker; September 10th 2006 at 07:13 PM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Sep 2006
    Posts
    134
    Thank you so much.. Only 3 more left..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Problem Solving.
    Posted in the Geometry Forum
    Replies: 4
    Last Post: August 15th 2010, 03:12 PM
  2. Problem Solving
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 5th 2010, 07:45 PM
  3. Solving for X problem?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: February 20th 2009, 10:27 AM
  4. Solving a Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: December 10th 2008, 08:27 PM
  5. Problem Solving
    Posted in the Algebra Forum
    Replies: 11
    Last Post: August 15th 2008, 06:42 PM

Search Tags


/mathhelpforum @mathhelpforum