I'm confused

[austral]

Consider two conditions, x²-3x-10<0 and |x-2|<a, on a real number x, where a is a positive real number.
i) A necessary and sufficient condition for x²-3x-10<0 is that (A)<x<(B)
ii) The range of values of a such that |x-2|<a is a necessary condition for x²-3x-10<0 is (C)
iii) The range of values of a such that |x-2|<a is a sufficient condition for x²-3x-10<0 is (D)
For each of A~D, choose the most appropriate expression from below:
i) 2, 5, -2, -5
ii) a≥2, a≥3, a≥4
iii) 0<a≤2, 0<a≤3, 0<a≤5

In the answer sheet, it says that A = -2, B = 5, C = a≥4 and D = 0<a≤3.

For i) I used -b/2a to find the vertex of x²-3x-10<0 (3/2, -19/4), then graphed the parabola and found that the X intercepts are -2 and 5. The graph ended up being gigantic and pretty annoying to draw, though. Is there a better way to find the X intercepts?

For ii), I first replaced the x in |x-2|<a with -2: |-2-2|<a, |-4|<a, 4<a. Then, I replaced it with 5: |5-2|<a, |3|<a, 3<a. So for |x-2|<a to be a necessary condition for x²-3x-10<0, a≥4.

For iii), I did the same as in ii), but for |x-2|<a to be a sufficient condition for x²-3x-10<0, a must be at least greater than or equal to 3. And the answer for iii) is 0<a≤3, but in 3<a we can see that 3 is less than a. So how can the answer be that a is less than or equal to 3 if 3 is less than a?

I'm confused. Maybe I'm doing everything wrong...

Thanks in advance.

 

[Dr.Peterson]

For i) I used -b/2a to find the vertex of x²-3x-10<0 (3/2, -19/4), then graphed the parabola and found that the X intercepts are -2 and 5. The graph ended up being gigantic and pretty annoying to draw, though. Is there a better way to find the X intercepts?

Have you learned about factoring? x²-3x-10 factors as (x-5)(x+2), which leads quickly to the result you need.

For ii), I first replaced the x in |x-2|<a with -2: |-2-2|<a, |-4|<a, 4<a. Then, I replaced it with 5: |5-2|<a, |3|<a, 3<a. So for |x-2|<a to be a necessary condition for x²-3x-10<0, a≥4.

We are looking for an interval about x=2 that will cover the interval (-2,5). Since -2 is 4 less than 2, and 5 is 3 more than 2, a has to be at least 4, the greater of the two distances. This is equivalent to your work.

For iii), I did the same as in ii), but for |x-2|<a to be a sufficient condition for x²-3x-10<0, a must be at least greater than or equal to 3. And the answer for iii) is 0<a≤3, but in 3<a we can see that 3 is less than a. So how can the answer be that a is less than or equal to 3 if 3 is less than a?

This time you want an interval about x=2 that is entirely within (-2,5); so a has to be no more than the smaller of the two distances, which was 3. Why did you say "a must be at least greater than or equal to 3"?

I thought of both of these visually, in terms of the graphs of the intervals. Your more algebraic approach can feel more formal, but to me it's more dangerous, as inequalities are easy to get backward. It appears that you focused too much on |x-2|<a and not enough on the fact that you want a sufficient condition.

 

[austral]

This time you want an interval about x=2

Why is the interval about x=2? Does that 2 come from |x-2|<a? I understand that if |x-2|<4, the solutions are -2 and 6, and that covers (-2, 5). But I don't get the greater and smaller distances from 2 part...

 

[Dr.Peterson]

Why is the interval about x=2? Does that 2 come from |x-2|<a? I understand that if |x-2|<4, the solutions are -2 and 6, and that covers (-2, 5). But I don't get the greater and smaller distances from 2 part...

Yes, |x-2|<a is the interval centered around 2, going a units above and below 2.