Results 1 to 4 of 4

Math Help - Inequality

  1. #1
    Member
    Joined
    Mar 2006
    Posts
    125

    Inequality

    Hi guys,

    for some reason this equation is giving me hard time. There shouldn't be a solution to it....and I do get one.

    My mistake is somewhere in the first line, because Maple makes the same mistake if I let it start from my second line of solution.

    My first line is to power each side, so I do ()^2 for each side, meaning that on the left side the square root is gone, and on the right side, I can something like (a-b)^2.

    Somehow it doesn't work, can someone solve it please ?
    Attached Thumbnails Attached Thumbnails Inequality-ineq.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    The inequality is equivalent to:

    \sqrt[]{1-2x}+2<\sqrt[]{2-2x}

    Taking squares and smplifying:

    4\;\sqrt[]{1-2x}<-3

    This is absurd because

    4\;\sqrt[]{1-2x}\;\geq 0

    Regards.

    ---
    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2006
    Posts
    125
    thank you,
    why did you move the "2" before you took square ?
    isn't it allowed to take square when the statement is like (a-b)^2 and not (a+b)^2 ?
    because if you do take square without moving it, you get an answer...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    Quote Originally Posted by WeeG View Post
    thank you,
    why did you move the "2" before you took square ?
    isn't it allowed to take square when the statement is like (a-b)^2 and not (a+b)^2 ?
    because if you do take square without moving it, you get an answer...
    I understand your question. First of all, when we write \sqrt{a} we are considering the positive root. It is not necessary to move the "2", but the solution is more complicated.

    The original inequality is:

    \sqrt{1-2x}<\sqrt{2-2x}-2 \qquad [1]

    Tahing into account that \sqrt{1-2x}\geq 0, necessary:

    \sqrt{2-2x}-2>0\qquad [2]

    If you solve [2], you will obtain:

    x<-1\qquad [3]

    If you take squares in [1] without moving the "2" you will obtalin:

    x\geq 7/32

    and this is a contradiction with [3].

    Regards.

    ---
    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2011, 08:20 PM
  2. Replies: 3
    Last Post: December 12th 2010, 01:16 PM
  3. inequality
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 26th 2010, 04:34 AM
  4. inequality
    Posted in the Algebra Forum
    Replies: 4
    Last Post: July 24th 2010, 12:08 PM
  5. Inequality
    Posted in the Algebra Forum
    Replies: 0
    Last Post: October 8th 2009, 03:06 AM

Search Tags


/mathhelpforum @mathhelpforum