Math / Algebra

Question3 4 5 6 7 8 9 10 e solutions to the inequality $(x-3)(x+5) \leq 0 ?$ $\begin{array}{l} x \leq 5\} \\ \leq x \leq-3\} \\ \leq x \leq 3\} \\ \leq x \leq 5\} \end{array}$

Studdy Solution

STEP 1

1. The inequality $(x-3)(x+5) \leq 0$ is a quadratic inequality.
2. The inequality involves finding the range of $x$ values that satisfy the condition.
3. The solutions will be intervals on the real number line.

STEP 2

1. Identify the critical points by solving the equation $(x-3)(x+5) = 0$ .
2. Determine the intervals defined by the critical points.
3. Test each interval to determine where the inequality holds true.
4. Write the solution set based on the intervals that satisfy the inequality.

STEP 3

Find the critical points by setting each factor equal to zero:
$x - 3 = 0 \quad \text{or} \quad x + 5 = 0$

STEP 4

Solve for $x$ :
$x = 3 \quad \text{or} \quad x = -5$

STEP 5

Determine the intervals defined by the critical points. The critical points divide the number line into the following intervals:
1. $(-\infty, -5)$
2. $[-5, 3]$
3. $(3, \infty)$

STEP 6

Test each interval to determine where the inequality $(x-3)(x+5) \leq 0$ holds:
- For $(-\infty, -5)$ , choose a test point like $x = -6$ : $(x-3)(x+5) = (-6-3)(-6+5) = (-9)(-1) = 9$ , which is not $\leq 0$ .
- For $[-5, 3]$ , choose a test point like $x = 0$ : $(x-3)(x+5) = (0-3)(0+5) = (-3)(5) = -15$ , which is $\leq 0$ .
- For $(3, \infty)$ , choose a test point like $x = 4$ : $(x-3)(x+5) = (4-3)(4+5) = (1)(9) = 9$ , which is not $\leq 0$ .

STEP 7

Based on the test results, write the solution set. The inequality holds for the interval:
$[-5, 3]$
The solution to the inequality $(x-3)(x+5) \leq 0$ is:
$\boxed{[-5, 3]}$

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Question3 4 5 6 7 8 9 10 e solutions to the inequality $(x-3)(x+5) \leq 0 ?$ $\begin{array}{l} x \leq 5\} \\ \leq x \leq-3\} \\ \leq x \leq 3\} \\ \leq x \leq 5\} \end{array}$

Studdy Solution

STEP 1

1. The inequality $(x-3)(x+5) \leq 0$ is a quadratic inequality.
2. The inequality involves finding the range of $x$ values that satisfy the condition.
3. The solutions will be intervals on the real number line.

STEP 2

1. Identify the critical points by solving the equation $(x-3)(x+5) = 0$ .
2. Determine the intervals defined by the critical points.
3. Test each interval to determine where the inequality holds true.
4. Write the solution set based on the intervals that satisfy the inequality.

STEP 3

Find the critical points by setting each factor equal to zero:
$x - 3 = 0 \quad \text{or} \quad x + 5 = 0$

STEP 4

Solve for $x$ :
$x = 3 \quad \text{or} \quad x = -5$

STEP 5

Determine the intervals defined by the critical points. The critical points divide the number line into the following intervals:
1. $(-\infty, -5)$
2. $[-5, 3]$
3. $(3, \infty)$

STEP 6

STEP 7

Based on the test results, write the solution set. The inequality holds for the interval:
$[-5, 3]$
The solution to the inequality $(x-3)(x+5) \leq 0$ is:
$\boxed{[-5, 3]}$

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Question3 4 5 6 7 8 9 10 e solutions to the inequality (x−3)(x+5)≤0?(x-3)(x+5) \leq 0 ?(x−3)(x+5)≤0? x≤5}≤x≤−3}≤x≤3}≤x≤5}\begin{array}{l} x \leq 5\} \\ \leq x \leq-3\} \\ \leq x \leq 3\} \\ \leq x \leq 5\} \end{array}x≤5}≤x≤−3}≤x≤3}≤x≤5}​

Studdy Solution

STEP 1

1. The inequality (x−3)(x+5)≤0(x-3)(x+5) \leq 0(x−3)(x+5)≤0 is a quadratic inequality.2. The inequality involves finding the range of xxx values that satisfy the condition.3. The solutions will be intervals on the real number line.

STEP 2

STEP 3

Find the critical points by setting each factor equal to zero:x−3=0orx+5=0 x - 3 = 0 \quad \text{or} \quad x + 5 = 0 x−3=0orx+5=0

STEP 4

Solve for x x x:x=3orx=−5 x = 3 \quad \text{or} \quad x = -5 x=3orx=−5

STEP 5

Determine the intervals defined by the critical points. The critical points divide the number line into the following intervals:1. (−∞,−5) (-\infty, -5) (−∞,−5)2. [−5,3] [-5, 3] [−5,3]3. (3,∞) (3, \infty) (3,∞)

STEP 6

STEP 7

Based on the test results, write the solution set. The inequality holds for the interval:[−5,3] [-5, 3] [−5,3] The solution to the inequality (x−3)(x+5)≤0(x-3)(x+5) \leq 0(x−3)(x+5)≤0 is:[−5,3] \boxed{[-5, 3]} [−5,3]​

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question3 4 5 6 7 8 9 10 e solutions to the inequality $(x-3)(x+5) \leq 0 ?$ $\begin{array}{l} x \leq 5\} \\ \leq x \leq-3\} \\ \leq x \leq 3\} \\ \leq x \leq 5\} \end{array}$

1. The inequality $(x-3)(x+5) \leq 0$ is a quadratic inequality.
2. The inequality involves finding the range of $x$ values that satisfy the condition.
3. The solutions will be intervals on the real number line.

Find the critical points by setting each factor equal to zero:
$x - 3 = 0 \quad \text{or} \quad x + 5 = 0$

Solve for $x$ :
$x = 3 \quad \text{or} \quad x = -5$

Determine the intervals defined by the critical points. The critical points divide the number line into the following intervals:
1. $(-\infty, -5)$
2. $[-5, 3]$
3. $(3, \infty)$

Based on the test results, write the solution set. The inequality holds for the interval:
$[-5, 3]$
The solution to the inequality $(x-3)(x+5) \leq 0$ is:
$\boxed{[-5, 3]}$