# How do I solve this rational inequality?

I'm having difficulty finding any help online for this equation:

$$\frac{3x^2+2x-5}{2x^2+5x}\le0$$

I need to solve the inequality, and then put it in interval notation.

So far I have taken the numerator and factored it to find the zeros, and I did the same with the denominator.

For the numerator I factored out $(3x+5)(x-1)$, and then I wrote that the $x$ values are $-\frac{5}{3}$ and 1 Then I factored the denominator and got $x(2x+5)$, and I put the $x$ values as $0$ and $-\frac{5}{2}$

What am I doing wrong? I tried graphing it on a calculator, but I just confused myself even more.

• You are trying to figure out when your expression is less or equal to zero. $-5/3$ and $1$ makes the expression equal to zero, but you are missing a bunch that makes it less than zero. – Braindead Jun 11 '15 at 5:04
• Also, $0$ and $-5/2$ makes the expression undefined. – Braindead Jun 11 '15 at 5:04
• Perhaps sign analysis is the topic you're looking for. Every step you've demonstrated above is correct so far. You're just missing the key step in solving these type of problems. – MathNewbie Jun 11 '15 at 5:06
• $\frac{3x^2+2x-5}{2x^2+5x}\le0\Leftrightarrow (3x^2+2x-5)(2x^2+5x)\le0$, $x\ne 0$, $x\ne -5/2$ – Michael Galuza Jun 11 '15 at 5:09
• @abel, no. If $P/Q \le 0$, then $P/Q\cdot Q^2 \le 0 \Leftrightarrow PQ \le 0$ – Michael Galuza Jun 11 '15 at 16:10

Solving the inequality

$$\frac{(3x+5)(x-1)}{x(2x+5)} \leq 0,$$

my 'standard' technique is to use a number line, test point, and multiplicities of zeroes.

$$-\underset{-5/2}{\bullet}-\underset{-5/3}{\circ}-\underset{0}{\circ}-\underset{1}{\bullet}\to\underset{x}{\ }$$

(Note here we're using filled in circles, $\bullet$, for $x$-values at which the function is defined (the zeros), and hollow circles, $\circ$, for the zeros of the denominator, where the function is undefined)

Picking, for example, $x = -1$ (the test point), your fraction is $\dfrac{2\cdot(-2)}{(-1)\cdot(3)}> 0$. Since the multiplicities of zeros of the numerator and denominator are all $1$, that is, odd, we know the sign of the function changes between each zero/undefined point. Knowing then that our function is positive on the interval $[-5/3, 0)$ since it's positive at $x = -1 \in [-5/3, 0)$ and switches signs at all special $x$-values, our sign chart looks like

$$\overset{+}{-}\underset{-5/2}{\bullet}\overset{-}{-}\underset{-5/3}{\circ}\overset{+}{-}\underset{0}{\circ}\overset{-}{-}\underset{1}{\bullet}\overset{+}{\longrightarrow}\underset{x}{\ }$$

and from there, we can read off the solution (paying attention to the zeros vs. undefined points), $x \in [-5/2, -5/3) \cup (0, 1]$.

Rather than using multiplicities of zeros, you could just pick a point in each interval between 'special' $x$-values and see if the function is positive or negative there. This takes more work, so using multiplicities is definitely to your advantage.

Edit for multiplicities:

Let's think of your function as one big product, and make a sign chart for each factor. \begin{array}{c|c|c|c|c|c|} &(-\infty, -5/2)&(-5/2, -5/3)&(-5/3, 0)&(0, 1)&(1, \infty) \\\hline 3x+5 &-&-&+&+&+ \\\hline x - 1 &-&-&-&-&+ \\\hline x &-&-&-&+&+ \\\hline 2x+5 &-&+&+&+&+\\\hline\\ \text{product}&+&-&+&-&+ \end{array}

How would things change if, for example, rather than just a factor of $(x - 1) = (x - 1)^1$, we had $(x - 1)^2$, so that $x = 1$ is a zero of multiplicity $2$?

\begin{array}{c|c|c|c|c|c|} &(-\infty, -5/2)&(-5/2, -5/3)&(-5/3, 0)&(0, 1)&(1, \infty) \\\hline 3x+5 &-&-&+&+&+ \\\hline (x - 1)^2 &+&+&+&+&+ \\\hline x &-&-&-&+&+ \\\hline 2x+5 &-&+&+&+&+\\\hline\\ \text{product}&-&+&-&+&+ \end{array}

Now, when $x = 1$ (corresponding to when we cross into the last column), our function doesn't change signs like it did before! Graphically, the function now "looks like" $y = x^2$ (shifted to the right) when we zoom in on $x = 1$, whereas it used to look more like $y = x$ (again shifted). The difference is that $y = x^2$ is positive when it isn't zero, unlike $y = x$ which changes signs at its zero. In a rough sense the function behaves like $y = x$ or $y = x^2$ at each zero of the numerator, depending on whether the multiplicity of that zero is odd or even.

I've skirted talking about how we're not really multiplying everything in the sign chart, but multiplying instead by the reciprocals of some factors. This doesn't change the sign, but you would compare the behavior near zeros of the denominator to functions like $y = 1/x$ (which changes signs) or $y = 1/x^2$ (which stays the same sign), to get a picture of the behavior near zeros of the denominator.

• The multiplicities do seem to make things so much easier, but I haven't been able to figure out how that works, yet. Do you know of any way I might be able to get a better understanding of the concept? – matryoshka Jun 11 '15 at 6:05
• @Grace I understand, it takes time for multiplicities to really "click". I've made an edit with more information about multiplicities, in the hope that it helps. – pjs36 Jun 11 '15 at 15:50

If either the numerator is less than zero, or the denominator is less than zero, but not both, then the fraction will be less than zero. You can find the intervals where this holds easily from your factorisations.

Then you need to consider boundary cases: when is the numerator zero (and the denominator non-zero)?

observe that all the critical points are single zeros. therefore the rational function switches sign across every one of them. knowing that the function has an asymptote $y = 3/2$ tells you that it is positive at $x = -\infty.$ now use the fact that rational function switches sign every critical point to determine the sign in each interval separated by the critical points.