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$$\frac{x(6+x)}{3+2x}\leq x$$ $$\frac{6x+x^2}{3+2x}-x\leq 0$$ $$(6x+x^2)(3+2x)-x(3+2x)^2\leq 0$$ $$x\neq\frac{-3}{2}$$

$$18x+12x^2+3x^2+2x^3-3x-2x^3=15x^2+15x\leq0$$ $$15x(x+1)\leq0$$ $$0\leq x \leq-1$$

Where I get it wrong? the answer is $x\geq3$ $-\frac{3}{2}\leq x\leq0$

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  • $\begingroup$ From the 2nd line, if you're multiplying through by $3+2x$ your equation should result in $6x+x^2-x(3+2x)^2 \leq 0$ $\endgroup$ Commented Dec 28, 2014 at 22:27
  • $\begingroup$ @MikeMiller I can not multiplying by $3+2x$ because I do not know if it is negative or not that why I multiple in $(3+2x)^2$ $\endgroup$
    – gbox
    Commented Dec 28, 2014 at 22:30
  • $\begingroup$ you still can't throw away $(3+2x)$ in going from the 3rd line to the 4th. $\endgroup$
    – abel
    Commented Dec 28, 2014 at 22:34
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    $\begingroup$ @gbox sorry, my first comment shouldn't be a square, but you can consider both cases afterwards. $\endgroup$ Commented Dec 28, 2014 at 22:40
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    $\begingroup$ i posted my explanation as an answer. check it out. $\endgroup$
    – abel
    Commented Dec 28, 2014 at 22:48

3 Answers 3

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Hint: Consider two cases: $x\geq0$ and $x<0$. Think also, when multiplying by $3+2x$ preserves the inequality.

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  • $\begingroup$ it is $-x^2+3x \leq 0$ and $-x^2+3x>0$? $\endgroup$
    – gbox
    Commented Dec 28, 2014 at 22:40
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    $\begingroup$ @gbox Well, in your solution two cases should be considered before obtaining the third line. But earlier division by $x$ simplifies the solution. $\endgroup$
    – Kola B.
    Commented Dec 28, 2014 at 22:56
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The error is on the 3rd line: you expanded $-x(3+2x)^2=-3x-2x^3$ instead of $-9x-12x^2-4x^3$. However, it's shorter to factor: \begin{align} p(x)&=(6x+x^2)(3+2x)-x(3+2x)^2 )\\ &= x(3+2x)(6+x-3-2x\\ &=x(3+2x)(3-x) \end{align} As $\lim_{x\to +\infty}p(x)=-\infty$, we deduce at once from the Intermediate Value Theorem that $p(x)\le 0$ for $x\ge 3$, then $p(x) \ge 0\,\,$ if $0\le x\le3$, $\,p(x)\le 0\,\,$ again if $-\frac{3}{2}\le x \le 0\,\,$ and finally $p(x)\ge 0\,\,$ if $x\le -\frac{3}{2}$.

So the solutions are $-\frac{3}{2}\le x \le 0$ or $x\ge 3$.

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here i will write out the solution here.

$\begin{eqnarray} {x(6+x) \over 3 + 2x} - x & \le & 0\\ {x(6+x) -x(3+2x) \over 3 + 2x} & \le & 0\\ {x^2 + 6x - 3x -2x^2 \over 3 + 2x} & \le & 0\\ {x(3-x) \over 3 + 2x} & \le & 0 \end{eqnarray}$

the critical numbers of the rational function are $0, 3$ and $-3/2$ by sign analysis , you can conclude that $$ {x(3-x) \over 3 + 2x} \le 0 \ on \ -3/2 < x \le 0 \ and \ 3 \le x < \infty $$

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  • $\begingroup$ by sign analysis you mean to check when $x(3-x)>0$ and $3+2x<0$ or $x(3-x)<0$ and $3+2x<0$? $\endgroup$
    – gbox
    Commented Dec 28, 2014 at 22:54
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    $\begingroup$ @gbox, i meant the rational function may switch(in this case, it does) signs only across the critical numbers. so all need to do is pick a test point in each of the four segments separated by the critical numbers and decide the sign for the whole segment. e.g. if you evaluate the rational function at $x = 1,$ you find the value $LHS = 2/5$ which is positive, so $LHS > 0, 0 < x < 3$. do you get what i mean by sign analysis? $\endgroup$
    – abel
    Commented Dec 28, 2014 at 23:01

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