Question$\frac{1}{x+2}+\frac{3}{x} \leqslant-2$

Studdy Solution
Now we need to solve these quadratic inequalities:
For Case 1: 2x^2 + 8x + 6 ≤ 0 The quadratic formula gives us roots at x = -3 and x = -1. The parabola opens upward (coefficient of x^2 is positive). So, the solution for this case is: -3 ≤ x ≤ -1
For Case 2: 2x^2 + 8x + 6 ≥ 0 This is the opposite of Case 1, so the solution is: x ≤ -3 or x ≥ -1
Now, we need to combine these results with the conditions from Step 2:
Case 1 is valid when x(x+2) > 0, which occurs when x < -2 or x > 0 Case 2 is valid when x(x+2) < 0, which occurs when -2 < x < 0
Combining all of this information:
For x < -2: We use Case 2, so x ≤ -3 is the solution. For -2 < x < 0: We use Case 2, but there are no solutions in this interval. For x > 0: We use Case 1, but there are no solutions in this interval.
Therefore, the final solution is:
$$x \leq -3$$
The solution to the inequality is:
$$\boxed{x \leq -3}$$