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Thread: Inequalities

  1. #1
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    Inequalities

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

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

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

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

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

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

    $\displaystyle x\geqslant 0$ and $\displaystyle x\leqslant 16$

    therefore, $\displaystyle 0\leqslant x\leqslant 16$

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

    So shouldn't the right answer be... $\displaystyle 0\leqslant x\leqslant 16, x\neq \frac{1}{9}$ ? why is this not the answer? Please advise, thanks
    Last edited by Punch; Jun 3rd 2011 at 02:47 AM.
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  2. #2
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    The difficulty is that your solution to the original inequality is wrong. There is no reason to say $\displaystyle x\ne \frac{1}{3}$. It is $\displaystyle x= -\frac{1}{3}$ that would cause the left side to not exist.
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  3. #3
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    sorry that was a typo, so why shouldn't the solution include $\displaystyle \sqrt{x}\neq-\frac{1}{3}$

    $\displaystyle x\neq \frac{1}{9}$
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  4. #4
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    $\displaystyle \sqrt{x}$ 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.
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  5. #5
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    Problem 1

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

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

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

    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 $\displaystyle x = -\frac{1}{3}$ because we cannot divide by 0, there would be no solution.

    So our answer is:

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

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

    Problem 2

    Now on your next problem:

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


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

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



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

    $\displaystyle \sqrt{x} = 4$

    $\displaystyle x = 16$



    $\displaystyle (\sqrt{x}+1) = 0$

    $\displaystyle \sqrt{x} = -1$

    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: $\displaystyle (3\sqrt{x}+1)^2 \neq 0$

    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 $\displaystyle \sqrt{x} \leq 4$

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

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

    Good luck you were on the right track it just looks like some typos threw you off.
    Last edited by Stro; Jun 3rd 2011 at 07:26 PM.
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  6. #6
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    $\displaystyle \displaystyle \frac{(x + 1)(4 - x)}{(3x + 1)^2} \geq 0$

    You first need to note that $\displaystyle \displaystyle x \neq - \frac{1}{3}$, 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

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

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

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

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

    $\displaystyle \displaystyle x \in \left[-1, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, 4\right]$
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  7. #7
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    Alternatively,
    the fraction is undefined for $\displaystyle x=-\frac{1}{3}$

    For all other x, the denominator is positive.

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

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

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

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

    Both are positive if $\displaystyle -1< x< 4$

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

    $\displaystyle -1\le\ x\le\ 4$
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