Question

if x ^ 2 - 5x + 6 / X + 4 is greater than equal to zero solve by a>=0 and b>0

Answer 1

x ∈ (-4, 2] ∪ [3, ∞)

Answer 2

To solve the inequality $\frac{x^2 - 5x + 6}{x + 4} \ge 0$, we follow these steps:

1. Factorize the numerator:
   The quadratic expression $x^2 - 5x + 6$ can be factored as $(x - 2)(x - 3)$.
   So, the inequality becomes:
   $\frac{(x - 2)(x - 3)}{x + 4} \ge 0$

2. Identify Critical Points:
   The critical points are the values of $x$ that make the numerator or the denominator equal to zero.

   • From the numerator: $x - 2 = 0 \implies x = 2$
     $x - 3 = 0 \implies x = 3$
   • From the denominator: $x + 4 = 0 \implies x = -4$

3. Plot Critical Points on a Number Line and Use the Wavy Curve Method:
   Arrange the critical points in ascending order on a number line: -4, 2, 3.

Now, we determine the sign of the expression $\frac{(x - 2)(x - 3)}{x + 4}$ in each interval. Since all factors $(x+4)$, $(x-2)$, and $(x-3)$ have an odd power (power 1), the sign of the expression will alternate as we move across each critical point.

*   **For $x > 3$ (e.g., $x=4$):**  

$(x-2) = (4-2) = 2$ (Positive)
$(x-3) = (4-3) = 1$ (Positive)
$(x+4) = (4+4) = 8$ (Positive)
Expression: $\frac{(+)(+)}{(+)} = (+)$

*   **For $2 < x < 3$ (e.g., $x=2.5$):**  

$(x-2) = (2.5-2) = 0.5$ (Positive)
$(x-3) = (2.5-3) = -0.5$ (Negative)
$(x+4) = (2.5+4) = 6.5$ (Positive)
Expression: $\frac{(+)(-)}{(+)} = (-)$

*   **For $-4 < x < 2$ (e.g., $x=0$):**  

$(x-2) = (0-2) = -2$ (Negative)
$(x-3) = (0-3) = -3$ (Negative)
$(x+4) = (0+4) = 4$ (Positive)
Expression: $\frac{(-)(-)}{(+)} = (+)$

*   **For $x < -4$ (e.g., $x=-5$):**  

$(x-2) = (-5-2) = -7$ (Negative)
$(x-3) = (-5-3) = -8$ (Negative)
$(x+4) = (-5+4) = -1$ (Negative)
Expression: $\frac{(-)(-)}{(-)} = (-)$

4. Determine the Solution Set:
   We need the intervals where the expression is $\ge 0$. This means where it is positive or equal to zero.
   • The expression is positive in the intervals $(-4, 2)$ and $(3, \infty)$.
   • The expression is zero when the numerator is zero, i.e., $x=2$ or $x=3$. These points are included in the solution.
   • The expression is undefined when the denominator is zero, i.e., $x=-4$. This point must be excluded.

Combining these, the solution set is:
$x \in (-4, 2] \cup [3, \infty)$

The instruction "solve by a>=0 and b>0" refers to one of the cases for $\frac{A}{B} \ge 0$. For a complete solution, we consider two cases: Case 1: $A \ge 0$ AND $B > 0$ $(x-2)(x-3) \ge 0 \implies x \in (-\infty, 2] \cup [3, \infty)$ $x+4 > 0 \implies x \in (-4, \infty)$ Intersection of these two: $x \in (-4, 2] \cup [3, \infty)$.

Case 2: $A \le 0$ AND $B < 0$ $(x-2)(x-3) \le 0 \implies x \in [2, 3]$ $x+4 < 0 \implies x \in (-\infty, -4)$ Intersection of these two: $\emptyset$ (empty set).

Combining both cases, the solution is $x \in (-4, 2] \cup [3, \infty)$.