# Thread: Getting the signs right with inequalities

1. ## Getting the signs right with inequalities

Hi folks,

I am getting the wrong range of values to the following question:

calculate the range of x for which:

$\left | \frac{2x + 5}{x^2 - 4} \right | \ge \frac{1}{5}$

the modulus means that both numerator and denominator are positive, therefore:

$2x + 5 \ge \frac{1}{5}(x^2 - 4)$

$-x^2 + 10x + 29 \ge 0$ ....................................(1)

$x = \frac{-10 \pm \sqrt{100 + 4.1.29}}{-2} \Rightarrow 5 \pm \sqrt{216}$

$x = 5 \pm 3\sqrt{6}$

$(x - (5 - 3\sqrt{6})) (x - (5 + 3\sqrt{6})) \ge 0$

$(x - 5 + 3\sqrt{6}) (x - 5 - 3\sqrt{6}) \ge 0$

for this inequality to be true:

$x - 5 + 3\sqrt{6} \ge 0 \ and \ x - 5 - 3\sqrt{6} \ge 0$

$x \ge 5 - 3\sqrt{6} \ and \ x \ge 5 + 3\sqrt{6}$

for both to be true simultaneously x must be $\ge 5 + 3\sqrt{6}$

or

$x - 5 + 3\sqrt{6} \le 0 \ and \ x - 5 - 3\sqrt{6} \le 0$

$x \le 5 - 3\sqrt{6} \ and \ x \le 5 + 3\sqrt{6}$

for both to be true simultaneously x must be $\le 5 - 3\sqrt{6}$

so the range is

$5 + 3\sqrt{6} \le x \le 5 - 3\sqrt{6}$

$5 - 3\sqrt{6} \le x \le 5 + 3\sqrt{6}$

I think the reason for the problem is that (1) can be framed as shown above or written as

$x^2 -10x -29 \le 0$ ....................................(2)

in both cases the roots are the same $x = 5 \pm 3\sqrt{6}$

so my question to the forum is should I use equation (1) or (2) ? (2) gives the right answer, but I should be able to use either. Is it the roots that should be different? I was a bit unsure when I was taking the negative of $\pm$ !

2. ## Re: Getting the signs right with inequalities

I do not agree with the given answer. Attached is the graph of the equivalent inequality

$\bigg| \dfrac{2x+5}{x^2-4} \bigg| -\dfrac{1}{5} \ge 0$

As depicted in the graph, the inequality is true over the intervals $[-7,-3]$, $[5-3\sqrt{6},-2)$, $(-2,2)$, $(2,5+3\sqrt{6}]$

3. ## Re: Getting the signs right with inequalities

Originally Posted by s_ingram

the modulus means that both numerator and denominator are positive, or both are negative, or the numerator is zero
.

4. ## Re: Getting the signs right with inequalities

The solutions of $\displaystyle |A|\geq B$ where $\displaystyle B$ is positive are given by the solutions of

$\displaystyle A\geq B$

together with (Union) the solutions of

$\displaystyle A\leq -B$

First, solve $\displaystyle \frac{2x+5}{x^2-4}\geq \frac{1}{5}$
solution:

$\displaystyle 5-3 \sqrt{6}\leq x<-2$ or $\displaystyle 2<x\leq 5+3 \sqrt{6}$

Next, solve $\displaystyle \frac{2x+5}{x^2-4}\leq - 1/5$
solution:

$\displaystyle -7\leq x\leq -3$ or $\displaystyle -2<x<2$

5. ## Re: Getting the signs right with inequalities

Good Idea (I guess everyone says that!)

Yes, it seems I am only doing part of the question. Let me give it a try!

6. ## Re: Getting the signs right with inequalities

Originally Posted by s_ingram
Not sure I understand this: the modulus sign makes a value positive so |-3| = 3, so the numerator and denominator are independently positive. If the numerator is zero, then the function would not be defined.
$\dfrac{0}{4} = 0$ is definitely defined.

7. ## Re: Getting the signs right with inequalities

Thanks Skeeter. The full answer given is $-7 \le x \le -3, (5 - 3\sqrt{6}) \le x \le (5 + 3\sqrt{6}) \ with \ x \ne \pm 2$

8. ## Re: Getting the signs right with inequalities

Originally Posted by s_ingram
Not sure I understand this: the modulus sign makes a value positive so |-3| = 3, so the numerator and denominator are independently positive. If the numerator is zero, then the function would not be defined.
absolute value of a rational function ...

$\bigg|\dfrac{f(x)}{g(x)}\bigg| = \dfrac{f(x)}{g(x)}$ if $\dfrac{f(x)}{g(x)} \ge 0$ ... $f(x) = 0$ or $f(x)$ and $g(x)$ have the same sign.

$\bigg|\dfrac{f(x)}{g(x)}\bigg| = -\dfrac{f(x)}{g(x)}$ if $\dfrac{f(x)}{g(x)} < 0$ ... $f(x)$ and $g(x)$ have opposite signs.

in either case, $g(x) \ne 0$ ... if the denominator equals zero, the function is undefined.

9. ## Re: Getting the signs right with inequalities

Hi SlipEternal,

I meant if $x = \pm 2$ then the modulus function is undefined.

10. ## Re: Getting the signs right with inequalities

Hi Skeeter,

I agree with everything you say in the above. I was using these methods to get to my answer. Thanks to the replies I have so far, I see that I was only doing part of the question, but even there I seem to be getting into a mess. The issue is that I get different answers, using the logic you just laid out, when I try to solve $x^2 -10x -29 \le 0$ and $-x^2 +10x +29 \ge 0$.

11. ## Re: Getting the signs right with inequalities

Hi Idea,

I followed your suggestion and it certainly helps!
When I solve $\frac{2x + 5}{x^2 - 4} \le \frac{-1}{5} \ I get \ -7 \le x \le -3$ which is great.
But I still have a problem with $\frac{2x + 5}{x^2 - 4} \ge \frac{1}{5}$ which gives me $(5 + 3\sqrt{6}) \le x \le (5 - 3\sqrt{6})$ instead of $(5 - 3\sqrt{6}) \le x \le (5 + 3\sqrt{6})$, using the reasoning shown in the original post. No one, so far, has explained where I am going wrong. Any ideas?

12. ## Re: Getting the signs right with inequalities

Originally Posted by s_ingram
Hi Idea,

I followed your suggestion and it certainly helps!
When I solve $\frac{2x + 5}{x^2 - 4} \le \frac{-1}{5} \ I get \ -7 \le x \le -3$ which is great.
But I still have a problem with $\frac{2x + 5}{x^2 - 4} \ge \frac{1}{5}$ which gives me $(5 + 3\sqrt{6}) \le x \le (5 - 3\sqrt{6})$ instead of $(5 - 3\sqrt{6}) \le x \le (5 + 3\sqrt{6})$, using the reasoning shown in the original post. No one, so far, has explained where I am going wrong. Any ideas?
Let us solve the inequality

$\displaystyle \frac{2x+5}{x^2-4}\leq \frac{-1}{5}$

$\displaystyle \frac{2x+5}{x^2-4}+ \frac{1}{5}\leq 0$

$\displaystyle \frac{x^2+10 x+21}{5 \left(x^2-4\right)}\leq 0$

$\displaystyle \frac{(x+7)(x+3)}{5 (x-2)(x+2)}\leq 0$

There are four factors on the left that affect the sign of the expression giving rise to 5 intervals on the real line

$\displaystyle (-\infty ,-7,-3,-2,2,+\infty )$

We check each of those 5 intervals to see which make the expression less or equal to zero

and we find not only $\displaystyle -7\leq x\leq -3$ but also $\displaystyle -2<x<2$

13. ## Re: Getting the signs right with inequalities

another method ...

note $|u| = \sqrt{u^2}$

$\left|\dfrac{2x+5}{x^2-4}\right| \ge \dfrac{1}{5}$

$\sqrt{\left(\dfrac{2x+5}{x^2-4}\right)^2} \ge \dfrac{1}{5}$

$\dfrac{(2x+5)^2}{(x^2-4)^2} \ge \dfrac{1}{25}$ ... keep in mind that $x \ne \pm 2$

$25(2x+5)^2 \ge (x^2-4)^2$

$25(2x+5)^2 - (x^2-4)^2 \ge 0$

$\left[5(2x+5) - (x^2-4)\right] \cdot \left[5(2x+5) + (x^2-4)\right] \ge 0$

$(29+10x-x^2)(x^2+10x+21) \ge 0$

$(29+10x-x^2)(x+7)(x+3) \ge 0$

critical values are $x=5\pm3\sqrt{6}$, $x=-7$, $x=-3$, and $x=\pm 2$

plot each critical value on a number line and "test" a value within each interval in the last inequality (see attached diagram) ...

14. ## Re: Getting the signs right with inequalities

Thanks skeeter. You put all the pieces together. There was certainly more to this than I thought! Many thanks.