# Solving inequalities with fractions with unfactorable polynomials

So I've been cracking my head open trying to solve this inequality:

$$\frac {x+1}{2-x} \le \frac {x}{3+x}$$

I've been taught you have to put all factors to one side of the inequality (leaving zero on the other side) and then factor the polynomials. So I did the following:

$$\frac {x+1}{2-x}-\frac {x}{3+x} \le 0$$

Then:

$$\frac {(x+1)(3+x)- x(2-x)}{(2-x)(3+x)}\le 0$$

Which equals to:

$$\frac {2x^2+2x+3}{(2-x)(3+x)}\le 0$$

But that's where I've got a problem. I've tried different ways but I can't seem to find a way to express $2x^2+2x+3$ as a product of two binomials, and (as far as I know) I need to do so to solve it following these steps: http://www.purplemath.com/modules/ineqrtnl.htm Can someone please help me out? Is there any way to factorize the polynomial or another way to solve this inequality?

This is my first post by the way, so sorry if I made any mistakes. English is also not my first language, sorry for that. Thanks a lot in advance.

• Since the discriminant is negative, the polynomial in the numerator is always positive. Alternatively, ou may write that polynomial as $$\frac{(2x+1)^2+5}2$$, which shows it is in fact always positive. Commented Mar 24, 2014 at 0:34
• You can also just draw a graph of $y = (x+1)/(2-x)$ and a graph of $y= x/(3+x)$ are see how they compare... Commented Mar 24, 2014 at 0:41

As chubakueno says, the determinant of the numerator is $2^2 - 4\cdot 2 \cdot 3 = -20 < 0$. So the numerator is always strictly positive.

So the sign of your fraction is determined by the sign of your denominator. For $x>2$ the denominator is negative (as the first term $2-x$ is negative; the second is positive). So the whole fraction is negative (a positive number divided by a negative one is negative).

For $x \in (-3,2)$ both terms in the denominator are positive; so the whole fraction is also positive (arguing as above).

For $x < -3$ the denominator is negative (the first term in the denominator is postive; the second is negative). The fraction is therefore negative.

For $x=2$ the question doesn't make sense; similarly for $x=-3$.

The inequality is therefore satisfied for $x \in (-\infty,-3) \cup (2,\infty)$.

EDIT Proof the the determinant being negative implies the quadratic is positive: remember, if the determinant of the quadratic $ax^2 +bx +c$ is negative, then the equation $ax^2 +bx +c = 0$ has no solutions. Assume further that$a>0$ as it is in our case. Then the shape of the quadratic is a "U"-shape (as opposed to an "n"-shape), and does not intersect the $x$-axis. It is therefore always positive.

• Thanks! I'm practically trying to re-learn algebra all over again and this was great. I remember that if a polynomial's determinant is negative, the expression is always positive for any value of x, but I can't remember why. Perhaps you could clarify that for me? Commented Mar 24, 2014 at 0:55
• That's a pleasure. See the edit above where I explain this in heuristic terms (as opposed to giving a completely formal proof - I think the heuristic version might be more useful to you). Commented Mar 24, 2014 at 1:01
• Wow, thanks a lot, man! This really helped me out. Commented Mar 24, 2014 at 1:13

Your calculations this far are correct. Now, determine the sign of each of the two polynomials that appear in the numerator and denominator of th fraction

1. $n(x)=2x^2+2x+3$. Calculate the discriminant $$\Delta=b^2-4ac=2^2-4(2)(3)=4-24=-20<0$$ so that the polynomial in the numerator is always positive. Thus it can be ignored from the subsequent analysis, since it does not affect the sign of the LHS.
2. $d(x)=(2-x)(3+x)$. It is immediate (from the form of the polynomial) that $$\begin{cases}d(x) > 0, & \text{for } x \in (-3,2) \\ d(x)\le 0, & \text{else }\end{cases}$$

Thus the inequality holds whenever $d(x)\le 0$, i.e. for $x \notin (-3,2)$, that is for $$x\in(-\infty, -3) \cup(2,+\infty)$$ The values $x=-3$ and $x=2$ are excluded because the nullify the denominator.

• This was great, thank you! Commented Mar 24, 2014 at 1:04