Use algebraic​ and/or graphical methods to solve the inequality.

Question

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.

Answer 1

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 \( (-\infty, -6] \cup [0, 6] \).