# Inequalities problem

1. Apr 25, 2016

### erisedk

1. The problem statement, all variables and given/known data
Find the set of all $x$ for which $\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}$

2. Relevant equations

3. The attempt at a solution

I'm getting two different sets of answers with two different methods:

Method 1-Wrong

$\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}$

$\dfrac{2x^2 + 5x + 2}{2x} < x + 1$

$\dfrac{2x^2 + 5x + 2}{2x} - (x+1) < 0$

$\dfrac{2x^2 + 5x + 2 - 2x(x+1)}{2x} < 0$

$\dfrac{3x + 2}{2x} < 0$

$x \in \left( \dfrac{-2}{3}, 0 \right)$

Method 2, the correct one

$\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}$

$\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0$

$\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0$

$\dfrac{3x + 2}{(2x + 1)(x + 1)(x + 2)} < 0$

$x \in (-2,-1) \cup \left(\dfrac{-2}{3} , \dfrac{-1}{2}\right)$

2. Apr 25, 2016

### SammyS

Staff Emeritus
In method 1, the first step is correct only if the left hand side and right hand side have the same sign.

If the left hand side is positive and the right hand side is negative, then which way does the inequality go after taking the reciprocal?

3. Apr 25, 2016

### erisedk

Oh! Got it, thank you :)

4. Apr 25, 2016

### Ray Vickson

How do you know you can go from $\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0$ to $\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0$? Certainly, if you multiply both sides of an inequality by a positive quantity, the inequality remains unchanged in direction. However, if you multiply both sides by a negative, the inequality is reversed.

5. Apr 25, 2016

### SammyS

Staff Emeritus
It looks more like OP combined rational expressions by using a common denominator, rather than by multiplying the whole expression by anything.

6. Apr 25, 2016

### Ray Vickson

Agreed, but I would have preferred that the OP answer the question.