Question

(2x^2+18x-4)/(x^2+9x+8)>2

Answer 1

Ответ:

как то так..................

Answer 2

To solve the inequality $\frac{2x^2 + 18x - 4}{x^2 + 9x + 8} > 2$, follow these steps:

1. Find the Critical Points:

   • First, find the critical points of the equation by setting the numerator equal to zero and solving for $x$: $2x^2 + 18x - 4 = 0$
   • Factor the quadratic equation or use the quadratic formula to find the critical points.

2. Determine the Sign of the Inequality:

   • Choose test points within the intervals created by the critical points (i.e., points less than the smallest critical point, points between the critical points, and points greater than the largest critical point).

3. Test the Intervals:

   • Plug in the test points into the original inequality to determine which intervals satisfy the inequality.

4. Express the Solution:

   • Write down the final solution in interval notation.

Let's proceed with the calculations:

1. Find the Critical Points: The equation $2x^2 + 18x - 4 = 0$ can be simplified by dividing by 2: $x^2 + 9x - 2 = 0$

   This can be factored as $(x+1)(x+8) = 0$, so the critical points are $x = -1$ and $x = -8$.

2. Determine the Sign of the Inequality: We now have three intervals to test: $(- \infty, -8)$, $(-8, -1)$, and $(-1, \infty)$.

3. Test the Intervals:

   • For $x < -8$, choose $x = -10$ (arbitrarily less than -8): $\frac{2(-10)^2 + 18(-10) - 4}{(-10)^2 + 9(-10) + 8} = \frac{220}{8} > 2$

   • For $-8 < x < -1$, choose $x = -5$ (arbitrarily between -8 and -1): $\frac{2(-5)^2 + 18(-5) - 4}{(-5)^2 + 9(-5) + 8} = \frac{-12}{8} < 2$

   • For $x > -1$, choose $x = 0$ (arbitrarily greater than -1): $\frac{2(0)^2 + 18(0) - 4}{(0)^2 + 9(0) + 8} = \frac{-4}{8} < 2$

4. Express the Solution:

   • The solution to the inequality is $x \in (- \infty, -8) \cup (-1, \infty)$.

Please note that we exclude the values $x = -1$ and $x = -8$ from the solution because those values make the denominator of the original expression zero, which is not allowed in mathematical operations.