Question

9х^2+3х-2 ————— ≥0 х+2

Answer 1

Ответ: решение смотри на фотографии

Объяснение:

Answer 2

To solve the inequality $\frac{9x^2 + 3x - 2}{x + 2} \geq 0$, we need to find the intervals of $x$ that satisfy the inequality. To do this, follow these steps:

Step 1: Find the critical points by setting the numerator equal to zero and solving for $x$:

$9x^2 + 3x - 2 = 0$

Step 2: Factor or use the quadratic formula to find the critical points. In this case, the quadratic equation does not factor easily, so let's use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 9$, $b = 3$, and $c = -2$.

$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 9 \cdot (-2)}}{2 \cdot 9}$ $x = \frac{-3 \pm \sqrt{9 + 72}}{18}$ $x = \frac{-3 \pm \sqrt{81}}{18}$ $x = \frac{-3 \pm 9}{18}$

So the critical points are $x = \frac{6}{18} = \frac{1}{3}$ and $x = \frac{-12}{18} = -\frac{2}{3}$.

Step 3: Determine the sign of the expression $\frac{9x^2 + 3x - 2}{x + 2}$ in the intervals separated by the critical points.

Choose a test point in each interval and plug it into the expression. Here are the intervals to consider: $(-\infty, -\frac{2}{3}), (-\frac{2}{3}, \frac{1}{3}),$ and $(\frac{1}{3}, \infty)$.

For $x = 0$:

$\frac{9(0)^2 + 3(0) - 2}{0 + 2} = \frac{-2}{2} = -1$

For $x = -1$:

$\frac{9(-1)^2 + 3(-1) - 2}{-1 + 2} = \frac{9 - 3 - 2}{1} = 4$

For $x = 1$:

$\frac{9(1)^2 + 3(1) - 2}{1 + 2} = \frac{9 + 3 - 2}{3} = \frac{10}{3}$

Step 4: Determine when the expression is greater than or equal to zero.

From the test points, we can see that the expression is positive (greater than zero) in the interval $x \in (-\frac{2}{3}, \frac{1}{3})$ and negative (less than zero) in the intervals $x \in (-\infty, -\frac{2}{3})$ and $x \in (\frac{1}{3}, \infty)$.

Step 5: Final solution

The solution to the inequality $\frac{9x^2 + 3x - 2}{x + 2} \geq 0$ is $x \in (-\frac{2}{3}, \frac{1}{3})$