Inequality

• Apr 9th 2011, 07:13 AM
aonin
Inequality
Hey guys I'm confused whether I doing these right, I used 2 different methods to solve this problem, and they seem to be giving different answers

$\text{solve} \dfrac{1}{x-3}> 4$

1st method:

$\dfrac{1}{x-3}> 4$

$\dfrac{1}{(x-3)^2}>16$

$1>16(x-3)^2$

$1>16x^2-96x+144$

$16x^2-96x+143<0$

$x=\dfrac{96\pm\sqrt{96^2-4(16)(143)}}{32}$

$\dfrac{88}{32}

2nd method

$\dfrac{1}{x-3}> 4$

$\dfrac{1}{x-3}-4>0$

$\text{multiply both sides by}\hspace{5mm}(x-3)^2$

$(x-3)-4(x-3)^2>0$

$(x-3)(1-4(x-3))>0$

$(x-3)(13-4x)>0$

$3

they give different answers...? Pls let me know if i've done anything wrong (Doh)

Thanks
• Apr 9th 2011, 07:22 AM
Prove It
Whenever you square an equation or inequation to aid in solving it, you are likely to bring in extraneous solutions.

What I would do is invert both sides, noting that inverting both sides of an inequation changes the direction of the inequality sign. The rest should be obvious.
• Apr 9th 2011, 07:23 AM
TheEmptySet
First we need to get a common denominator on both sides

$\displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}$

simplify to get

$\displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}$

So the two critical numbers(where the inequality can chage) are $\displaystyle x=3,\quad x=\frac{13}{4}$

If you test some points you will get

$\displaystyle \left(3,\frac{13}{4}\right)$
• Apr 9th 2011, 07:28 AM
Prove It
Quote:

Originally Posted by TheEmptySet
First we need to get a common denominator on both sides

$\displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}$

simplify to get

$\displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}$

So the two critical numbers(where the inequality can chage) are $\displaystyle x=3,quad x=\frac{13}{4}$

If you test some points you will get

$\displaystyle (-\infty,3) \cup \left(\frac{13}{4},\infty \right)$

Sorry but for any $\displaystyle x$ in $\displaystyle (-\infty, 3)$, $\displaystyle \frac{1}{x - 3}$ gives something negative, which is clearly NOT greater than $\displaystyle 4$...
• Apr 9th 2011, 07:35 AM
HallsofIvy
The way I would have done it is this: $\frac{1}{x-3}> 4$ is the same as $0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $(-\infty, 3)$, $(3, 13/4)$, and $(13/4, \infty)$. If x= 0, which is less than 3, $\frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\frac{13}{4}= 3\frac{1}{4}$, $\frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.

Therefore, the inequality is true on the interval $\left(3, \frac{13}{4}\right)$.
• Apr 9th 2011, 07:35 AM
TheEmptySet
Quote:

Originally Posted by Prove It
Sorry but for any $\displaystyle x$ in $\displaystyle (-\infty, 3)$, $\displaystyle \frac{1}{x - 3}$ gives something negative, which is clearly NOT greater than $\displaystyle 4$...

You are correct I cant add this morning! thanks
• Apr 9th 2011, 07:41 AM
veileen
"If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.

The second method is wrong. You don't know if $\frac{(x-3)^2}{x-3}=x-3$ or $3-x$.

First method is wrong too. You can square only after you say that x must be bigger then 3.
• Apr 9th 2011, 07:47 AM
Prove It
Quote:

Originally Posted by HallsofIvy
The way I would have done it is this: $\frac{1}{x-3}> 4$ is the same as $0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $(-\infty, 3)$, $(3, 13/4)$, and $(13/4, \infty)$. If x= 0, which is less than 3, $\frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\frac{13}{4}= 3\frac{1}{4}$, $\frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.

Therefore, the inequality is true on the interval $\left(3, \frac{13}{4}\right)$.

Actually $\displaystyle \frac{1}{x - 3} > 4$ is the same as $\displaystyle 0 > 4 - \frac{1}{x - 3}$...
• Apr 9th 2011, 08:16 AM
veileen
A way this kind of inequalities can be solved:

$\frac{1}{x-3}>4 \Leftrightarrow 4-\frac{1}{x-3}<0 \Leftrightarrow \frac{4x-13}{x-3}<0$

http://img193.imageshack.us/img193/5701/frlu.jpg

You want that fraction be negativ, so $x \in \left ( 3, \frac{13}{4} \right )$.
• Aug 14th 2011, 04:37 AM
aonin
Re: Inequality
Quote:

Originally Posted by veileen
The second method is wrong. You don't know if $\frac{(x-3)^2}{x-3}=x-3$ or $3-x$.

sorry for bringing up an old thread, but I thought about this inequality thing again, looking over what you wrote isn't $\frac{(x-3)^2}{x-3}=x-3$ always?, isn't this true for all real numbers x, for example sub in x=5 then the answer would be 2, which is 5-3, if subbing in lets say a negative number -1, then it would be 16/-4=-1-3 which is true again, the same is true for x=0 or a positive number smaller than 3, eg 1
• Aug 14th 2011, 04:53 AM
Siron
Re: Inequality
The statement:
$\frac{(x-3)^2}{x-3}=3-x$ is not true because, that would mean $(3-x)\cdot(x-3)=(x-3)^2 \Leftrightarrow -x^2+6x-9=x^2-6x+9$ which is not true for all $x$.
• Aug 14th 2011, 01:23 PM
HallsofIvy
Re: Inequality
Quote:

Originally Posted by veileen
"If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.

There are a lot of things that anybody can do- that doesn't mean it is wrong.

The only places an inequality can change form ">" to "<" is where the function is 0 or is not continuous. For fractions that is where the numerator is 0 or where the denominator is 0. Once you have found those points, checking one point in each interval tells you which. What TheEmptySet said was perfectly valid.