Math Problem Statement

−(x − 3)(x + 5) < 0

Solution

We are tasked with solving the inequality:

(x3)(x+5)<0-(x - 3)(x + 5) < 0

Step 1: Eliminate the negative sign

Distribute the negative sign into the inequality:

(x3)(x+5)=(1)(x3)(x+5)-(x - 3)(x + 5) = (-1)(x - 3)(x + 5)

This becomes:

(x3)(x+5)>0(x - 3)(x + 5) > 0

Now, we are solving the inequality:

(x3)(x+5)>0(x - 3)(x + 5) > 0

Step 2: Find the critical points

Set each factor to zero to find the critical points:

x3=0x=3x - 3 = 0 \quad \Rightarrow \quad x = 3 x+5=0x=5x + 5 = 0 \quad \Rightarrow \quad x = -5

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

  1. (,5)(-\infty, -5)
  2. (5,3)(-5, 3)
  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 (,5)(-\infty, -5), say x=6x = -6:

(x3)(x+5)=(63)(6+5)=(9)(1)=9>0(x - 3)(x + 5) = (-6 - 3)(-6 + 5) = (-9)(-1) = 9 > 0

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

(x3)(x+5)=(03)(0+5)=(3)(5)=15<0(x - 3)(x + 5) = (0 - 3)(0 + 5) = (-3)(5) = -15 < 0

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

(x3)(x+5)=(43)(4+5)=(1)(9)=9>0(x - 3)(x + 5) = (4 - 3)(4 + 5) = (1)(9) = 9 > 0

Step 4: Conclusion

The inequality holds where (x3)(x+5)>0(x - 3)(x + 5) > 0, which is in the intervals:

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

Thus, the solution to the inequality is:

x(,5)(3,)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 (x2)(x+4)>0(x - 2)(x + 4) > 0.
  2. Solve (x1)(x7)<0(x - 1)(x - 7) < 0.
  3. Solve (x+3)(x5)0(x + 3)(x - 5) \geq 0.
  4. Solve 2(x6)(x+1)0-2(x - 6)(x + 1) \leq 0.
  5. Solve (x+2)2(x4)<0(x + 2)^2(x - 4) < 0.

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

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Math Problem Analysis

Mathematical Concepts

Inequalities
Quadratic Expressions
Critical Points
Intervals

Formulas

Factoring and solving quadratic expressions
Inequality testing through interval analysis

Theorems

Sign analysis of products in intervals
Critical points theorem for inequalities

Suitable Grade Level

Grades 9-11