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Use algebraic and/or graphical methods to solve the inequality. 49 x squared minus x Superscript 4 Baseline greater than or eq...Question
Use algebraic and/or graphical methods to solve the inequality.
36 x squared minus x Superscript 4 Baseline greater than or equals 0
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Part 1
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution(s) is/are xequals
enter your response here.
(Use a comma to separate answers as needed.)
B.
The solutions are the values of x in the interval
enter your response here.
(Type your answer in interval notation.)
C.
There is no solution.
36 x squared minus x Superscript 4 Baseline greater than or equals 0
Question content area bottom
Part 1
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution(s) is/are xequals
enter your response here.
(Use a comma to separate answers as needed.)
B.
The solutions are the values of x in the interval
enter your response here.
(Type your answer in interval notation.)
C.
There is no solution.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the inequality \( 36x^2 - x^4 \geq 0 \), we first rearrange it to make factoring easier. Rewrite it as:
\[ -x^4 + 36x^2 \geq 0 \]
Factoring out \( -1 \) gives us:
\[ -(x^4 - 36x^2) \geq 0 \]
Now we can factor the expression inside the parentheses:
\[ -(x^2(x^2 - 36)) \geq 0 \]
This factors further into:
\[ -x^2(x - 6)(x + 6) \geq 0 \]
Next, we can find the critical points by setting each factor to zero:
1. \( -x^2 = 0 \) gives \( x = 0 \) (double root).
2. \( x - 6 = 0 \) gives \( x = 6 \).
3. \( x + 6 = 0 \) gives \( x = -6 \).
The critical points are \( x = -6, 0, 6 \). These points divide the number line into intervals:
1. \( (-\infty, -6) \)
2. \( (-6, 0) \)
3. \( (0, 6) \)
4. \( (6, \infty) \)
We will test each of these intervals to determine where the expression is non-negative.
1. **Interval \( (-\infty, -6) \):** Choose \( x = -7 \)
\[ -(-7)^2(-7 - 6)(-7 + 6) = -49(-13)(-1) = -49 \cdot 13 \cdot 1 > 0 \]
2. **Interval \( (-6, 0) \):** Choose \( x = -1 \)
\[ -(-1)^2(-1 - 6)(-1 + 6) = -1(-7)(5) = 35 > 0 \]
3. **Interval \( (0, 6) \):** Choose \( x = 1 \)
\[ -(1)^2(1 - 6)(1 + 6) = -1(-5)(7) = 35 > 0 \]
4. **Interval \( (6, \infty) \):** Choose \( x = 7 \)
\[ -(7)^2(7 - 6)(7 + 6) = -49(1)(13) = -637 < 0 \]
Now, we summarize the results:
- The expression is non-negative in the intervals \( (-\infty, -6) \), \( (-6, 0) \), and \( (0, 6) \).
- At the critical points, \( f(-6) = 0 \), \( f(0) = 0 \), and \( f(6) = 0 \).
Thus, in interval notation, the solution to \( 36x^2 - x^4 \geq 0 \) is:
\[
(-\infty, -6] \cup [0, 6]
\]
### Final Answer:
**B. The solutions are the values of \( x \) in the interval \( (-\infty, -6] \cup [0, 6] \).**
\[ -x^4 + 36x^2 \geq 0 \]
Factoring out \( -1 \) gives us:
\[ -(x^4 - 36x^2) \geq 0 \]
Now we can factor the expression inside the parentheses:
\[ -(x^2(x^2 - 36)) \geq 0 \]
This factors further into:
\[ -x^2(x - 6)(x + 6) \geq 0 \]
Next, we can find the critical points by setting each factor to zero:
1. \( -x^2 = 0 \) gives \( x = 0 \) (double root).
2. \( x - 6 = 0 \) gives \( x = 6 \).
3. \( x + 6 = 0 \) gives \( x = -6 \).
The critical points are \( x = -6, 0, 6 \). These points divide the number line into intervals:
1. \( (-\infty, -6) \)
2. \( (-6, 0) \)
3. \( (0, 6) \)
4. \( (6, \infty) \)
We will test each of these intervals to determine where the expression is non-negative.
1. **Interval \( (-\infty, -6) \):** Choose \( x = -7 \)
\[ -(-7)^2(-7 - 6)(-7 + 6) = -49(-13)(-1) = -49 \cdot 13 \cdot 1 > 0 \]
2. **Interval \( (-6, 0) \):** Choose \( x = -1 \)
\[ -(-1)^2(-1 - 6)(-1 + 6) = -1(-7)(5) = 35 > 0 \]
3. **Interval \( (0, 6) \):** Choose \( x = 1 \)
\[ -(1)^2(1 - 6)(1 + 6) = -1(-5)(7) = 35 > 0 \]
4. **Interval \( (6, \infty) \):** Choose \( x = 7 \)
\[ -(7)^2(7 - 6)(7 + 6) = -49(1)(13) = -637 < 0 \]
Now, we summarize the results:
- The expression is non-negative in the intervals \( (-\infty, -6) \), \( (-6, 0) \), and \( (0, 6) \).
- At the critical points, \( f(-6) = 0 \), \( f(0) = 0 \), and \( f(6) = 0 \).
Thus, in interval notation, the solution to \( 36x^2 - x^4 \geq 0 \) is:
\[
(-\infty, -6] \cup [0, 6]
\]
### Final Answer:
**B. The solutions are the values of \( x \) in the interval \( (-\infty, -6] \cup [0, 6] \).**
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