Question

$\frac{2x}{x^2+5x+2} \geq \frac{1}{x+1}$

Answer 1

$x\in\left(\frac{-5-\sqrt{17}}{2},\,-1\right)\cup\left[\frac{3-\sqrt{17}}{2},\,\frac{-5+\sqrt{17}}{2}\right)\cup\left[\frac{3+\sqrt{17}}{2},\,\infty\right)$

Answer 2

We wish to solve

$$\frac{2x}{x^2+5x+2} \ge \frac{1}{x+1}.$$

Step 1. Bring all terms to one side

$$\frac{2x}{x^2+5x+2} - \frac{1}{x+1} \ge 0.$$

Step 2. Write as a single rational expression

The common denominator is $(x^2+5x+2)(x+1)$ (note that $x+1\ne0$ and $x^2+5x+2\ne0$). Thus,

$$\frac{2x(x+1) - (x^2+5x+2)}{(x^2+5x+2)(x+1)} \ge 0.$$

Simplify the numerator:

$$2x(x+1) - (x^2+5x+2)= 2x^2+2x -x^2-5x-2 = x^2-3x-2.$$

So the inequality becomes

$$\frac{x^2-3x-2}{(x^2+5x+2)(x+1)} \ge 0.$$

Step 3. Find the critical points

1. Zeros of the numerator: Solve

   $$x^2-3x-2=0\quad\Longrightarrow\quad x=\frac{3\pm\sqrt{9+8}}{2}=\frac{3\pm\sqrt{17}}{2}.$$

   Let

   $$r_1=\frac{3-\sqrt{17}}{2}\quad \text{and}\quad r_2=\frac{3+\sqrt{17}}{2}.$$

   Numerically, $r_1\approx -0.56$ and $r_2\approx 3.56$.

2. Zeros of the factor $x+1$: $x=-1$.

3. Zeros of $x^2+5x+2$: Solve

   $$x^2+5x+2=0\quad\Longrightarrow\quad x=\frac{-5\pm\sqrt{25-8}}{2} = \frac{-5\pm\sqrt{17}}{2}.$$

   Let

   $$p_1=\frac{-5-\sqrt{17}}{2}\approx -4.56,\quad p_2=\frac{-5+\sqrt{17}}{2}\approx -0.44.$$

Remember that the expression is undefined when the denominator is zero, i.e. at $x=-1$, $x=p_1$ and $x=p_2$.

Step 4. Determine the sign in the various intervals

The critical points (in increasing order) are:

$$p_1\approx -4.56,\quad -1,\quad r_1\approx -0.56,\quad p_2\approx -0.44,\quad r_2\approx 3.56.$$

Thus, the real line is divided into these intervals:

1. $x<p_1$
2. $(p_1, -1)$
3. $(-1, r_1)$
4. $(r_1, p_2)$
5. $(p_2, r_2)$
6. $x>r_2$

Using test points in each interval (details omitted for brevity), one finds that the expression is nonnegative (i.e. positive or 0) in the intervals:

• $(p_1, -1)$
• $(r_1, p_2)$ (note that at $x=r_1$ the numerator is zero and it is allowed provided the denominator is nonzero)
• $(r_2, \infty)$ (with equality at $x=r_2$).

Step 5. Write the final answer, taking care to exclude points where the denominator is 0

Thus the solution set is:

x\in\left(\frac{-5-\sqrt{17}}{2},\,-1\right)\cup\left[\frac{3-\sqrt{17}}{2},\,\frac{-5+\sqrt{17}}{2}\right)\cup\left[\frac{3+\sqrt{17}}{2},\,\infty\right) $$. **Summary** - **Core Explanation:** 1. Bring all terms to one side. 2. Combine into a single rational expression. 3. Find zeros of the numerator and denominator. 4. Use a sign chart to find intervals where the expression is nonnegative. 5. Exclude values that make the denominator 0. - **Final Answer:** $$ x\in\left(\frac{-5-\sqrt{17}}{2},\,-1\right)\cup\left[\frac{3-\sqrt{17}}{2},\,\frac{-5+\sqrt{17}}{2}\right)\cup\left[\frac{3+\sqrt{17}}{2},\,\infty\right) $$