Results 1 to 7 of 7

Math Help - Inequalities

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Inequalities

    First part of the question says: Solve the inequality \frac{(x+1)(4-x)}{(3x+1)^2}\geqslant 0

    And after solving, the answer is -1\leqslant x\leqslant 4, x\neq \frac{1}{3}

    Then the next part of the question says: Hence deduce the solution to the inequality \frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2}\geqslant 0

    On comparing, i replaced x in the answer to the first part with \sqrt{x}

    so, -1\leqslant \sqrt{x}\leqslant 4, \sqrt{x}\neq -\frac{1}{3}

    -1\leqslant \sqrt{x} and \sqrt{x}\leqslant 4

    x\geqslant 0 and x\leqslant 16

    therefore, 0\leqslant x\leqslant 16

    Yes, that's right. The answer is correct but shouldn't the answer x\neq -\frac{1}{3} also be dealt with?

    So shouldn't the right answer be... 0\leqslant x\leqslant 16, x\neq \frac{1}{9} ? why is this not the answer? Please advise, thanks
    Last edited by Punch; June 3rd 2011 at 03:47 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,251
    Thanks
    1795
    The difficulty is that your solution to the original inequality is wrong. There is no reason to say x\ne \frac{1}{3}. It is x= -\frac{1}{3} that would cause the left side to not exist.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    sorry that was a typo, so why shouldn't the solution include \sqrt{x}\neq-\frac{1}{3}

    x\neq \frac{1}{9}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,251
    Thanks
    1795
    \sqrt{x} is never negative so there no need to exclude -1/3. And, if x= 1/9, then [tex]\sqrt{x}= 1/3[/itex], not -1/3.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2011
    Posts
    34
    Problem 1

    \frac{(x+1)(4-x)}{(3x+1)^2} \geq 0

    Let's change the equation up a bit to really look at it. All we'll do is pull a -1 out of the (4-x) to give -(x-4), then rearrange it:

    \frac{-(x+1)(x-4)}{(3x+1)^2} \geq 0

    So this is what we already know by looking at it:
    1. We have a negative degree two polynomial expression on top
    2. This expression has two real 0s so at some point x will be > 0
    2. We can easily figure out our zeros now for the equation (-1 and 4)
    3. Also it looks like if x = -\frac{1}{3} because we cannot divide by 0, there would be no solution.

    So our answer is:

    -1 \leq  x \leq 4 and x \neq -\frac{1}{3}

    Note another way to write this solution:
    -1 \leq  x  < -\frac{1}{3} and -\frac{1}{3} <  x \leq 4

    Problem 2

    Now on your next problem:

    \frac{(\sqrt{x}+1)(4-\sqrt{x})}{(3\sqrt{x}+1)^2} \geq 0


    \frac{-(\sqrt{x}+1)(\sqrt{x}-4)}{(3\sqrt{x}+1)^2} \geq 0

    It's going to be mostly the same except for two differences:



    1. Solving for our roots won't be as easy, for example:
    (\sqrt{x} - 4) = 0

    \sqrt{x} = 4

    x = 16



    (\sqrt{x}+1) = 0

    \sqrt{x} = -1

    There is no solution here, the sqrt(x) will never be negative.

    No solution, so only 16 for a root.

    So that was a little different but not too bad.

    2. There will never be a real value for x that would create a 0 in the bottom of your expression on the right

    For real numbers in x: (3\sqrt{x}+1)^2 \neq  0

    So we don't have to worry about a "no solution" problem on the bottom.


    And we understand that sqrt{x} will always remain non-negative. To explain this better, if we put a negative number in for x, we get a complex number in return. If we put a 0 in for x, we get a 0. And if we put a positive number in for x, we get a positive in return. Never a negative result.


    We can form our answer now to be \sqrt{x} \leq 4

    Or in other words, because \sqrt{x} is non-negative: 0 \leq \sqrt{x} \leq 4

    And thusly we have our solution: 0 \leq x \leq 16

    Good luck you were on the right track it just looks like some typos threw you off.
    Last edited by Stro; June 3rd 2011 at 08:26 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,804
    Thanks
    1576
    \displaystyle \frac{(x + 1)(4 - x)}{(3x + 1)^2} \geq 0

    You first need to note that \displaystyle x \neq - \frac{1}{3}, because that would give a zero denominator.
    Since the denominator is always nonnegative, you can multiply both sides by the denominator without changing the inequality sign

    \displaystyle (x + 1)(4 - x) \geq 0

    Quadratic inequalities are nearly always easiest solved by completing the square.

    \displaystyle \begin{align*} (x + 1)(4 - x) &\geq 0 \\ -x^2 + 3x + 4 &\geq 0 \\ -(x^2 - 3x - 4) &\geq 0 \\ -\left[x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 4\right] &\geq 0 \\ -\left[\left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{16}{4}\right] &\geq 0 \\ -\left[\left(x - \frac{3}{2}\right)^2 - \frac{25}{4}\right] &\geq 0 \\ -\left(x - \frac{3}{2}\right)^2 + \frac{25}{4} &\geq 0 \\ -\left(x - \frac{3}{2}\right)^2 &\geq -\frac{25}{4} \\ \left(x - \frac{3}{2}\right)^2 &\leq \frac{25}{4} \\ \left|x - \frac{3}{2}\right|^2 &\leq \frac{5}{2} \\ -\frac{5}{2} \leq x - \frac{3}{2} &\leq \frac{5}{2} \\ -1 \leq x &\leq 4\end{align*}

    Now, remembering that \displaystyle x \neq -\frac{1}{3}, that means our solution is

    \displaystyle x \in \left[-1, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, 4\right]
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Alternatively,
    the fraction is undefined for x=-\frac{1}{3}

    For all other x, the denominator is positive.

    Then for x\ne\ -\frac{1}{3}

    the solution reduces to (x+1)(4-x)\ge \0

    which will be true if both factors are negative or both are positive, or either is zero.

    Both are negative if x< -1 and x>4 at the same time, which is not possible.

    Both are positive if -1< x< 4

    Therefore, both factors cannot be negative and so for x\ne\ -\frac{1}{3}

    -1\le\ x\le\ 4
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. inequalities
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 20th 2011, 05:56 AM
  2. inequalities
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 20th 2011, 03:42 AM
  3. Inequalities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 1st 2009, 05:35 AM
  4. Inequalities
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 15th 2009, 12:26 PM
  5. inequalities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 14th 2008, 10:57 PM

Search Tags


/mathhelpforum @mathhelpforum