(Note: if you are curious about the formula, it is simplified from d = d0 + v0t + ½a0t2 , where d0=20, v0=0, and a0=−9.81 the acceleration due to gravity.)
OK, let's go.
First, let us sketch the question:
The distance we want is from 10 m to 15 m:
10 < d < 15
And we know the formula for d:
10 < 20 − 5t2 < 15
Now let's solve it!
First, let's subtract 20 from both sides:
−10 < −5t2 <−5
Now multiply both sides by −(1/5). But because we are multiplying by a negative number, the inequalities will change direction ... read Solving Inequalities to see why.
2 > t2 > 1
To be neat, the smaller number should be on the left, and the larger on the right. So let's swap them over (and make sure the inequalities still point correctly):
Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.
Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.
If a test point satisfies the original inequality, then the region that contains that test point is part of the solution.
Represent the solution in graphic form and in solution set form.
Example 1
Solve ( x – 3)( x + 2) > 0.
Solve ( x – 3)( x + 2) = 0. By the zero product property,
Make the boundary points. Here, the boundary points are open circles because the original inequality does not include equality (see Figure 1).
Select points from the different regions created (see Figure 2).
See whether the test points satisfy the original inequality.
Since x = –3 satisfies the original inequality, the region x < –2 is part of the solution. Since x = 0 does not satisfy the original inequality, the region –2 < x < 3 is not part of the solution. Since x = 4 satisfies the original inequality, the region x > 3 is part of the solution.
Represent the solution in graphic form and in solution set form. The graphic form is shown in Figure 3.
The solution set form is { x| x < –2 or x > 3}.
Figure 1. Boundary points.
Figure 2. Three regions are created.
Figure 3. Solution to Example
Example 2
Solve 9 x 2 – 2 ≤ –3 x.
By factoring,
Mark the boundary points using solid circles, as shown in Figure 4, since the original inequality includes equality.
Select points from the regions created (see Figure 5).
See whether the test points satisfy the original inequality.
Since x = –1 does not satisfy the original inequality, the region
Represent the solution in graphic form and in solution set form. The graphic form is shown in Figure 6.
The set form is
Figure 4. Solid dots mean inclusion.
Figure 5. Regions to test for Example
Figure 6. Solution to Example.
Example 3
Solve 4 t 2 – 9 < –4 t.
Since this quadratic is not easily factorable, the quadratic formula is used to solve it.
Reduce by dividing out the common factor of 4.
Since
Mark the boundary points using open circles, as shown in Figure 7, since the original inequality does not include equality.
Select points from the different regions created (see Figure 8).
See whether the test points satisfy the original inequality.
Since t = 3 does not satisfy the original inequality, the region
Represent the solution in graphic form and in solution set form. The graphic form is shown in Figure 9.
The solution set form is
Figure 7. Open dots mean exclusion.
Figure 8. Regions to test for Example.
Figure 9. Solution to Example.
Example 4
Solve
Since this quadratic is not factorable using rational numbers, the quadratic formula will be used to solve it.
These are imaginary answers and cannot be graphed on a real number line. Therefore, the inequality x 2 + 2 x + 5 < 0 has no real solutions.