# Thread: [SOLVED] Inequalities

1. ## [SOLVED] Inequalities

This is a little embarassing, but it seems like I do not understand why the answers are what they are for these inequalities.

1. $3x^2 - 2x-1 \geq 0$

First, I factor the left-hand side.

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

So, $x - 1 \geq 0$ if we divide both sides by '3x+1'.

Thus, $x \geq 1$

Here's the part I don't understand. Naturally, I think that if we divide 'x-1' from both sides. We would get

$3x+1 \geq 0$

which yields

$3x \geq -1 \therefore x \geq -\frac{1}{3}$ (*)

But this can not be right. By counterexample, let x = 0; this inequality is false.

It should be $x \leq -\frac{1}{3}$.

I believe the reason for this is since for (x-1), if x is a negative number, then (x-1) < 0. So in order for the inequality to be equal then '3x+1' must also be negative (since a negative times a negative equals a positive) which explains why it should be

$3x + 1 \leq 0$ and not $3x + 1 \geq 0$

Hence we solve, we should get $x \leq 1/3$. What am I doing wrong at (*)?

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2. Another confusing inequality.

$\frac{1}{3-x} < -2$

I multiply both sides by '3-x'
$1 < -2(3-x)$
$1 < -6+2x$
$7 < 2x$
$\frac{7}{2} < x$ (*)

This answer makes no sense. If x = 4, then the inequality is false.

Then, I tried a different way. I add both sides by 2

$\frac{1}{3-x} + 2 < 0$
$\frac{2(3-x)+1}{3-x} < 0$
$\frac{6-2x+1}{3-x} < 0$
$7 - 2x < 0$

So, $x < \frac{7}{2}$. Since, x can not be 3, or else the denominator is zero. I conclude that $3 < x < \frac{7}{2}$

What am I doing wrong at (*)?

Thank you.

2. The problem is when you divide both sides by a variable factor! you are losing information about the ineqality.

After you factor you get

$(3x+1)(x-1) \ge 0$ from here take your two zero's (this is where the inequality can change signs)

so we get $x=-\frac{1}{3} \mbox{ or } x=1$

Now we have three different parts of the number line

$\left(-\infty,-\frac{1}{3} \right)$

$\left(-\frac{1}{3},1 \right)$ and

$(1,\infty)$

Now we check a point from each interval to see if it is a solution

$\left(-\infty,-\frac{1}{3} \right)$ check $x=-1$

In the original and we get $(3(-1)+1)(-1-1)=4 \ge 0$ so the above set is a solution.

Now we check $\left(-\frac{1}{3},1 \right)$ with $x=0$ and we get $(3(0)+1)(0-1)=-1 \le 0$ so it is false and is not a solution.

finally we check $(1,\infty)$ with x=2 $(3(2)+1)(2-1)=7 \ge 0$ so this is a solution.

So the solution to the inequality is

$\left(-\infty,-\frac{1}{3} \right] \cup [1,\infty)$

For the rational ineqality get a common denomiantor and get the zero's from both the numerator and denomiator and do the same thing.

Good luck.

3. Thank you TheEmptySet. I was reviewing through my Calculus book and did not know/forgotten how to approach these problems. It was driving me crazy for a while. Your explanation makes more sense. So, I just choose arbitrary numbers within these intervals and check whether the inequality holds... ok. Thanks again. Cheers.