Answer:
Let's solve and graph these inequalities one by one:
1. x^2 >= 16:
To solve this inequality, we'll first find its roots:
x^2 - 16 = 0
(x + 4)(x - 4) = 0
The critical points are x = -4 and x = 4. Now, we'll test the regions between these points to determine the solution:
- When x < -4 or x > 4, x^2 is greater than 16.
- When -4 <= x <= 4, x^2 is less than or equal to 16.
Graph:
On a number line, mark -4 and 4 as open circles and shade both sides of the number line.
2. x^2 - 2x - 11 < 0:
To solve this quadratic inequality, first find its roots:
x^2 - 2x - 11 = 0
(x - 4)(x + 2) = 0
The critical points are x = 4 and x = -2. Now, we'll test the regions between these points to determine the solution:
- When x < -2 or x > 4, x^2 - 2x - 11 is less than 0.
- When -2 < x < 4, x^2 - 2x - 11 is greater than 0.
On a number line, mark -2 and 4 as open circles and shade the regions between them.
3. x^2 - 6x - 7 >= 0:
To solve this quadratic inequality, find its roots:
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
The critical points are x = 7 and x = -1. Now, we'll test the regions to determine the solution:
- When x <= -1 or x >= 7, x^2 - 6x - 7 is greater than or equal to 0.
- When -1 <= x <= 7, x^2 - 6x - 7 is less than 0.
On a number line, mark -1 and 7 as open circles and shade the regions outside these points.
4. x^2 - 3x >= 0:
x^2 - 3x = 0
x(x - 3) = 0
The critical points are x = 0 and x = 3. Now, we'll test the regions to determine the solution:
- When x <= 0 or x >= 3, x^2 - 3x is greater than or equal to 0.
- When 0 <= x <= 3, x^2 - 3x is less than 0.
On a number line, mark 0 and 3 as open circles and shade the regions outside these points.
5. 2x^2 - x >= 15:
Rearrange the inequality:
2x^2 - x - 15 >= 0
(2x + 5)(x - 3) >= 0
The critical points are x = -5/2 and x = 3. Now, we'll test the regions to determine the solution:
- When x <= -5/2 or x >= 3, 2x^2 - x - 15 is greater than or equal to 0.
- When -5/2 <= x <= 3, 2x^2 - x - 15 is less than 0.
On a number line, mark -5/2 and 3 as open circles and shade the regions outside these points.
These are the solutions and corresponding number line graphs for each of the given inequalities.
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Answers & Comments
Answer:
Let's solve and graph these inequalities one by one:
1. x^2 >= 16:
To solve this inequality, we'll first find its roots:
x^2 - 16 = 0
(x + 4)(x - 4) = 0
The critical points are x = -4 and x = 4. Now, we'll test the regions between these points to determine the solution:
- When x < -4 or x > 4, x^2 is greater than 16.
- When -4 <= x <= 4, x^2 is less than or equal to 16.
Graph:
On a number line, mark -4 and 4 as open circles and shade both sides of the number line.
2. x^2 - 2x - 11 < 0:
To solve this quadratic inequality, first find its roots:
x^2 - 2x - 11 = 0
(x - 4)(x + 2) = 0
The critical points are x = 4 and x = -2. Now, we'll test the regions between these points to determine the solution:
- When x < -2 or x > 4, x^2 - 2x - 11 is less than 0.
- When -2 < x < 4, x^2 - 2x - 11 is greater than 0.
Graph:
On a number line, mark -2 and 4 as open circles and shade the regions between them.
3. x^2 - 6x - 7 >= 0:
To solve this quadratic inequality, find its roots:
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
The critical points are x = 7 and x = -1. Now, we'll test the regions to determine the solution:
- When x <= -1 or x >= 7, x^2 - 6x - 7 is greater than or equal to 0.
- When -1 <= x <= 7, x^2 - 6x - 7 is less than 0.
Graph:
On a number line, mark -1 and 7 as open circles and shade the regions outside these points.
4. x^2 - 3x >= 0:
To solve this quadratic inequality, find its roots:
x^2 - 3x = 0
x(x - 3) = 0
The critical points are x = 0 and x = 3. Now, we'll test the regions to determine the solution:
- When x <= 0 or x >= 3, x^2 - 3x is greater than or equal to 0.
- When 0 <= x <= 3, x^2 - 3x is less than 0.
Graph:
On a number line, mark 0 and 3 as open circles and shade the regions outside these points.
5. 2x^2 - x >= 15:
Rearrange the inequality:
2x^2 - x - 15 >= 0
(2x + 5)(x - 3) >= 0
The critical points are x = -5/2 and x = 3. Now, we'll test the regions to determine the solution:
- When x <= -5/2 or x >= 3, 2x^2 - x - 15 is greater than or equal to 0.
- When -5/2 <= x <= 3, 2x^2 - x - 15 is less than 0.
Graph:
On a number line, mark -5/2 and 3 as open circles and shade the regions outside these points.
These are the solutions and corresponding number line graphs for each of the given inequalities.