Results 1 to 4 of 4

Math Help - Radical inequality

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    539

    Radical inequality

    Solve in R :
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by dhiab View Post
    Solve in R :
    Hi

    The derivative of the function f(x) = \sqrt{3-x}-\sqrt{x+1}-\frac12 is negative therefore f is decreasing.

    The problem is now to solve \sqrt{3-x}-\sqrt{x+1}-\frac12 = 0 which gives x = 1-\frac{\sqrt{31}}{8}.

    Thus \sqrt{3-x}-\sqrt{x+1}>\frac12 over \left[-1 ; 1-\frac{\sqrt{31}}{8}\right[
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744
    Hello, dhiab!

    What a bizarre problem . . .


    Solve in \mathbb{R}\!:\;\;\sqrt{3-x} - \sqrt{x+1} \:>\:\tfrac{1}{2}
    We can see that: . -1 \:\leq \:x\:\leq \:3

    Yet when I solved the inequality algebraically, I got misleading results.


    We have: . 2\sqrt{3-x} \:> \:2\sqrt{x+1} + 1

    Square: . 4(3-x) \:>\:4(x+1) + 4\sqrt{x+1} + 1

    Simplify: . 7 - 8x \:>\:4\sqrt{x+1}

    Square: . 49 - 112x + 64x^2 \:>\:16(x+1)

    Simplify: . 64x^2 - 128x + 33 \:>\:0

    This is an upward-opening parabola.
    . . When is it above the x-axis?
    Find its x-intercepts: . 64x^2 - 128x + 33 \:=\:0

    Quadratic Formula: . x \;=\;\frac{128 \pm\sqrt{7936}}{128} \;=\;\frac{128\pm16\sqrt{31}}{128} \;=\;\frac{8\pm\sqrt{31}}{8}

    The parabola is positive for: . \left(x < \frac{8-\sqrt{31}}{8}\right) \;\cup\; \left(x > \frac{8+\sqrt{31}}{8}\right)

    I realize that Squaring can produce extraneous roots,
    . . but "half" of this result is inapplicable.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    \text{We have: }\;\;\underbrace{\sqrt{3-x}}_{f(x)} \:- \:\underbrace{\sqrt{x+1}}_{g(x)} \:>\:\tfrac{1}{2}


    f(x)\!:\;\;y \:=\:\sqrt{3-x} \quad\Rightarrow\quad x \:=\:3 - y^2

    This is the upper half of a left-opening parabola with vertex (3,0).


    g(x)\!:\;\;y \:=\:\sqrt{x+1} \quad\Rightarrow\quad x \:=\:y^2 - 1

    This is the upper half of a right-opening parabola with vertex (-1,0).


    The question: When is f(x) more than a half-unit above g(x) ?
    Code:
                      |
            *   .     |                 *
                ::::*.|         *
                ::::::|:  *
                ::::::*       *
                :::*  |          *
                :*    |            *
          - - - * - - + - - - - - - * - - - - -
              -1 .    |            . 3
                   .  |          .
                      .       .
                      |   .
                    . |         .
            .         |                 .
                      |

    The solution is the shaded region: . -1 \leq\:x \:< \:\frac{8-\sqrt{31}}{8}


    Edit: Ha! . . . running-gag neat me to it!
    .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,237
    Thanks
    1795
    Blasted arithmetic mistakes!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Ring isomorphism maps a Jacobson radical into a Jacobson radical?
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: June 21st 2010, 11:45 AM
  2. Radical inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 12th 2009, 03:01 PM
  3. Radical-Radical equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: June 27th 2009, 04:38 AM
  4. Inequality cubic-radical
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 19th 2009, 05:13 PM
  5. Radical inside a radical?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 14th 2007, 10:59 PM

Search Tags


/mathhelpforum @mathhelpforum