1. Aug 5, 2008

### fr33pl4gu3

If one plots the graph of
y = x2-10 x+24

which is a parabola, one observes that it is above the x axis, i.e. y > 0, for two intervals of the real line. (The union of these intervals constitute the set of solutions of the quadratic inequality at the beginning of this question.)

Find the intervals.

How do you do this anyway??

Is it like take x = 3, and put it in and come out with y = 3, the the intervals would be (3,3)

2. Aug 5, 2008

### Varnick

So you're trying to find the range of x values that satisfy the equality y>0? If you set x^2 - 10x + 24 > 0 and solve normally, you should be fine. Is this correct?

V

3. Aug 5, 2008

### fr33pl4gu3

Yes, but i already solve that solution, but here is another one, that is similar to the previous one.

The solution set of x values of the ineqality
|-8 x-2| >= 10

may be expressed as the union of a pair of intervals of the real line. What are those intervals?

how to solve this??

I keep on getting the wrong answer.

4. Aug 5, 2008

### HallsofIvy

The best way to solve a problem like this is to first solve the equality. That is exactly what you did in the first problem.

That is, first solve |-8x- 2|= 10 which is the same as solving -8x- 2= 10 and -8x- 2= -10 which is the same as solving 8x- 2= 10. The two values where that is equal to 10 separate where it is <10 and > 10. Check one point in each of the three intervals to see which is which.

5. Aug 5, 2008

### moemoney

For the inequality | -8x - 2| = 10, its best to divide it into 2 cases; for all x values such that -8x - 2 > 0 (a) and all x values such that -8x - 2 < 0 (b). This step eliminates the use of absolute value functions.

For all x (a), the absolute value function would dissapear (because -8x - 2 will always
be greater than 0 for all x (a) values).

For all x (b), you would replace the absolute signs wiht a negative out infront (because -8x - 2 will always be less then 0).

So you would have -8x - 2 > 10 and -(-8x - 2) > 10 for two of the cases. (and these 2 cases would give you the union of pair intervals).

NOTE!!!!: You must go back and check your domains to see if your values are consitent with the orignal problem. This is a common mistake.