Question
Use algebraic and/or graphical methods to solve the inequality.
49 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 solutions are the values of x in the interval
enter your response here.
(Type your answer in interval notation.)
B.
The solution(s) is/are xequals
enter your response here.
(Use a comma to separate answers as needed.)
C.
There is no solution.
49 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 solutions are the values of x in the interval
enter your response here.
(Type your answer in interval notation.)
B.
The solution(s) is/are xequals
enter your response here.
(Use a comma to separate answers as needed.)
C.
There is no solution.
Answers
Answered by
GPT-4o mini
To solve the inequality \(49x^2 - x^4 \geq 0\), we can start by rearranging it:
\[
-x^4 + 49x^2 \geq 0
\]
This can be rewritten as:
\[
x^4 - 49x^2 \leq 0
\]
Now, we can factor the left side:
\[
x^2(x^2 - 49) \leq 0
\]
Next, we can further factor \(x^2 - 49\) using the difference of squares:
\[
x^2(x - 7)(x + 7) \leq 0
\]
Now we will find the critical points by setting each factor to zero:
1. \(x^2 = 0 \Rightarrow x = 0\)
2. \(x - 7 = 0 \Rightarrow x = 7\)
3. \(x + 7 = 0 \Rightarrow x = -7\)
Thus, the critical points are \(x = -7\), \(x = 0\), and \(x = 7\). These points divide the real number line into intervals:
1. \((- \infty, -7)\)
2. \((-7, 0)\)
3. \((0, 7)\)
4. \((7, \infty)\)
Next, we can test points from each interval to see where the expression \(x^2(x - 7)(x + 7)\) is less than or equal to zero.
1. Test a point in \((- \infty, -7)\), e.g. \(x = -8\):
\[
(-8)^2(-8 - 7)(-8 + 7) = 64(-15)(-1) = 960 > 0
\]
2. Test a point in \((-7, 0)\), e.g. \(x = -1\):
\[
(-1)^2(-1 - 7)(-1 + 7) = 1(-8)(6) = -48 < 0
\]
3. Test a point in \((0, 7)\), e.g. \(x = 1\):
\[
(1)^2(1 - 7)(1 + 7) = 1(-6)(8) = -48 < 0
\]
4. Test a point in \((7, \infty)\), e.g. \(x = 8\):
\[
(8)^2(8 - 7)(8 + 7) = 64(1)(15) = 960 > 0
\]
Now we summarize the test results:
- The expression is non-positive in the intervals \((-7, 0)\) and \((0, 7)\).
- We also include the points \(x = -7\), \(x = 0\), and \(x = 7\) since the original inequality is non-strict (i.e., \(\geq 0\)).
Thus, the solution in interval notation is:
\[
[-7, 0] \cup [0, 7]
\]
### Final Answer:
**A.** The solutions are the values of x in the interval \([-7, 0] \cup [0, 7]\).
\[
-x^4 + 49x^2 \geq 0
\]
This can be rewritten as:
\[
x^4 - 49x^2 \leq 0
\]
Now, we can factor the left side:
\[
x^2(x^2 - 49) \leq 0
\]
Next, we can further factor \(x^2 - 49\) using the difference of squares:
\[
x^2(x - 7)(x + 7) \leq 0
\]
Now we will find the critical points by setting each factor to zero:
1. \(x^2 = 0 \Rightarrow x = 0\)
2. \(x - 7 = 0 \Rightarrow x = 7\)
3. \(x + 7 = 0 \Rightarrow x = -7\)
Thus, the critical points are \(x = -7\), \(x = 0\), and \(x = 7\). These points divide the real number line into intervals:
1. \((- \infty, -7)\)
2. \((-7, 0)\)
3. \((0, 7)\)
4. \((7, \infty)\)
Next, we can test points from each interval to see where the expression \(x^2(x - 7)(x + 7)\) is less than or equal to zero.
1. Test a point in \((- \infty, -7)\), e.g. \(x = -8\):
\[
(-8)^2(-8 - 7)(-8 + 7) = 64(-15)(-1) = 960 > 0
\]
2. Test a point in \((-7, 0)\), e.g. \(x = -1\):
\[
(-1)^2(-1 - 7)(-1 + 7) = 1(-8)(6) = -48 < 0
\]
3. Test a point in \((0, 7)\), e.g. \(x = 1\):
\[
(1)^2(1 - 7)(1 + 7) = 1(-6)(8) = -48 < 0
\]
4. Test a point in \((7, \infty)\), e.g. \(x = 8\):
\[
(8)^2(8 - 7)(8 + 7) = 64(1)(15) = 960 > 0
\]
Now we summarize the test results:
- The expression is non-positive in the intervals \((-7, 0)\) and \((0, 7)\).
- We also include the points \(x = -7\), \(x = 0\), and \(x = 7\) since the original inequality is non-strict (i.e., \(\geq 0\)).
Thus, the solution in interval notation is:
\[
[-7, 0] \cup [0, 7]
\]
### Final Answer:
**A.** The solutions are the values of x in the interval \([-7, 0] \cup [0, 7]\).
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