Let's solve the inequality:

\[
-2x^2 - 5x + 9 \leq 6
\]

### Step 1: Move all terms to one side of the inequality.
Subtract 6 from both sides:

\[
-2x^2 - 5x + 9 - 6 \leq 0
\]

Simplifying:

\[
-2x^2 - 5x + 3 \leq 0
\]

### Step 2: Solve the corresponding equation.
We now solve:

\[
-2x^2 - 5x + 3 = 0
\]

Use the quadratic formula:

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

For the equation \( -2x^2 - 5x + 3 = 0 \), \( a = -2 \), \( b = -5 \), and \( c = 3 \).

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-2)(3)}}{2(-2)}
\]

Simplifying the discriminant:

\[
x = \frac{5 \pm \sqrt{25 + 24}}{-4}
\]
\[
x = \frac{5 \pm \sqrt{49}}{-4}
\]
\[
x = \frac{5 \pm 7}{-4}
\]

So, the two solutions are:

\[
x_1 = \frac{5 + 7}{-4} = \frac{12}{-4} = -3
\]
\[
x_2 = \frac{5 - 7}{-4} = \frac{-2}{-4} = \frac{1}{2}
\]

### Step 3: Analyze the sign of the quadratic expression.

The critical points are \( x = -3 \) and \( x = \frac{1}{2} \). We need to determine where \( -2x^2 - 5x + 3 \leq 0 \).

Test intervals around the critical points: \( (-\infty, -3) \), \( (-3, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).

1. **Interval \( (-\infty, -3) \):** Choose \( x = -4 \):

\[
-2(-4)^2 - 5(-4) + 3 = -2(16) + 20 + 3 = -32 + 20 + 3 = -9 \quad (\text{negative})
\]

2. **Interval \( (-3, \frac{1}{2}) \):** Choose \( x = 0 \):

\[
-2(0)^2 - 5(0) + 3 = 3 \quad (\text{positive})
\]

3. **Interval \( (\frac{1}{2}, \infty) \):** Choose \( x = 1 \):

\[
-2(1)^2 - 5(1) + 3 = -2 - 5 + 3 = -4 \quad (\text{negative})
\]

### Step 4: Conclusion.

The quadratic expression \( -2x^2 - 5x + 3 \leq 0 \) is negative or zero in the intervals \( (-\infty, -3] \) and \( \left[\frac{1}{2}, \infty \right) \).

Thus, the solution to the inequality is:

\[
x \in (-\infty, -3] \cup \left[\frac{1}{2}, \infty \right)
\]

---

Would you like further details on any step or have additional questions?

Here are 5 related questions to expand your understanding:
1. How do we determine where a quadratic expression is positive or negative?
2. What is the geometric interpretation of the solution for a quadratic inequality?
3. How does the quadratic formula help in solving inequalities?
4. What happens if the discriminant of a quadratic equation is negative?
5. How would the solution change if the inequality was strict (i.e., \(<\) instead of \( \leq \))?

**Tip:** Always sketch the graph of the quadratic to visually understand where it is above or below the x-axis!

Discover how to solve the quadratic inequality -2x² - 5x + 9 ≤ 6 with a comprehensive step-by-step explanation. The solution explores key algebraic concepts, including the quadratic formula, inequality solving, and critical points analysis. Ideal for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation