We are tasked with solving the inequality:

\[
-(x - 3)(x + 5) < 0
\]

### Step 1: Eliminate the negative sign
Distribute the negative sign into the inequality:

\[
-(x - 3)(x + 5) = (-1)(x - 3)(x + 5)
\]

This becomes:

\[
(x - 3)(x + 5) > 0
\]

Now, we are solving the inequality:

\[
(x - 3)(x + 5) > 0
\]

### Step 2: Find the critical points
Set each factor to zero to find the critical points:

\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]

These critical points, \( x = -5 \) and \( x = 3 \), divide the number line into three intervals:

1. \( (-\infty, -5) \)
2. \( (-5, 3) \)
3. \( (3, \infty) \)

### Step 3: Test the intervals
We will test points from each interval to determine where the inequality holds true.

- Test point in \( (-\infty, -5) \), say \( x = -6 \):

\[
(x - 3)(x + 5) = (-6 - 3)(-6 + 5) = (-9)(-1) = 9 > 0
\]

- Test point in \( (-5, 3) \), say \( x = 0 \):

\[
(x - 3)(x + 5) = (0 - 3)(0 + 5) = (-3)(5) = -15 < 0
\]

- Test point in \( (3, \infty) \), say \( x = 4 \):

\[
(x - 3)(x + 5) = (4 - 3)(4 + 5) = (1)(9) = 9 > 0
\]

### Step 4: Conclusion
The inequality holds where \( (x - 3)(x + 5) > 0 \), which is in the intervals:

\[
x \in (-\infty, -5) \cup (3, \infty)
\]

Thus, the solution to the inequality is:

\[
x \in (-\infty, -5) \cup (3, \infty)
\]

### Do you need more details or have any questions?

Here are 5 related questions for practice:

1. Solve \( (x - 2)(x + 4) > 0 \).
2. Solve \( (x - 1)(x - 7) < 0 \).
3. Solve \( (x + 3)(x - 5) \geq 0 \).
4. Solve \( -2(x - 6)(x + 1) \leq 0 \).
5. Solve \( (x + 2)^2(x - 4) < 0 \).

**Tip**: When solving inequalities, always test points from each interval to determine where the inequality holds.

Learn how to solve the inequality −(x − 3)(x + 5) < 0 using interval testing and critical points analysis. This step-by-step solution covers key algebraic techniques and is suitable for students in Grades 9-11.

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