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April 25th, 2014, 07:56 PM   #1
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Question Question on solving an inequality

In the inequality:

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

Why does the following manipulation lead to the wrong solution set?

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

$\displaystyle (x + 3)(x - 1) \geq 0$

$\displaystyle x \leq -3, x \geq 1$

Apparently, the solution set is $\displaystyle -3 \leq x \leq 0, x\geq 1$.

I don't understand how to get the correct solution.
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April 25th, 2014, 08:47 PM   #2
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Originally Posted by Mr Davis 97 View Post
In the inequality:

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

Why does the following manipulation lead to the wrong solution set?

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

$\displaystyle (x + 3)(x - 1) \geq 0$

$\displaystyle x \leq -3, x \geq 1$

Apparently, the solution set is $\displaystyle -3 \leq x \leq 0, x\geq 1$.

I don't understand how to get the correct solution.
No!$\displaystyle x\neq 0$.Make the table of changes for $\displaystyle x+3$,for $\displaystyle x-1$ and for $\displaystyle x$...........
Solution:
$\displaystyle x\in [-3,0)\cup [+1,+\infty)$

Last edited by Dacu; April 25th, 2014 at 08:59 PM.
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April 25th, 2014, 08:48 PM   #3
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The problem comes when you multiply by $x$. If $x<0$, then mutiplying by it will reverse the inequality.

\begin{align*}
\frac{(x + 3)(x - 1)}{x} &\geq 0 \\
(x + 3)(x - 1) &\leq 0 &\forall x \lt 0 \\
\end{align*}

And $(u + 3)(u - 1) \leq 0 \Longrightarrow -3 \leq u \leq 1$, but since we are limited to $x \lt 0$, we have $-3 \leq x \lt 0$.
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April 26th, 2014, 04:43 AM   #4
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When solving inequalities, always be careful about multiplication and division. Adding and subtracting terms is always okay, but when you multiply or divide you have to make sure you’re only doing it with positive quantities so as to preserve the direction of the inequality.

To find the solution set for $\displaystyle \frac{(x+3)(x-1)}x \geqslant 0$ one way is to divide the real line into into four intervals, $x < -3$, $-3 < x < 0$, $0 < x < 1$, $x > 1$, and see whether the factors are positive or negative (for the moment we’ll ignore the end points of the intervals):

$$\begin{array}{c|c|c|c|c}
& x+3 & x-1 & x & \frac{(x+3)(x-1)}x
\\ \hline \\ x < -3 & \mathrm{-ve} & \mathrm{-ve} & \mathrm{-ve} & \mathrm{-ve}
\\ \hline \\ -3 < x < 0 & \mathrm{+ve} & \mathrm{-ve} & \mathrm{-ve} & \mathrm{+ve}
\\ \hline \\ 0 < x < 1 & \mathrm{+ve} & \mathrm{+ve} & \mathrm{-ve} & \mathrm{-ve}
\\ \hline \\ x > 1 & \mathrm{+ve} & \mathrm{+ve} & \mathrm{+ve} & \mathrm{+ve}
\end{array}$$

Finally we note that $\displaystyle \frac{(x+3)(x-1)}x = 0$ when $x=-3$ or $x=1$ so we can add these points. This gives the solution set as $[-3,\,0]\cup[1,\,\infty)$.
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Last edited by Olinguito; April 26th, 2014 at 05:27 AM.
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