## anonymous 5 years ago 2x^2+x<15 polynomial inequality how do i solve and graph on a real number line?

1. anonymous

You can solve this by partitioning the interval first. Solve,$2x^2+x=15$ (i.e. set it to an equality) and solve for x. These values will partition the interval. You then take 'test points' in each of the intervals to see if, in that interval, the inequality holds. If it does at that point, it will for the entire interval. You test for all partitions like this.

2. anonymous

x=5/2 and x=3? just plot them?

3. anonymous

x=5/2 and x=-3.

4. anonymous

Yes, take a number line and plot those points. You'll see that it's partitioned into three regions.

5. anonymous

Now pick a point in the first region that's easy to do the math with. So, maybe pick x=-5.

6. anonymous

And test in the original inequality.

7. anonymous

right. i ended up shading the region between the two points becasue it includes all of them?

8. anonymous

At x=-5, 2x^2+x = 45, which is NOT less than 15, so that entire interval is out (i.e. everything to the left of -3).

9. anonymous

No...don't shade...it's just a visual to assist.

10. anonymous

The next interval contains 0, so use that. Then the left-hand side is 0, which IS less than 15, so this interval contains x's that solve the inequality.

11. anonymous

You then check the third region, say at x=10. Then LHS = 210 which is NOT less than 15, so everything above 5/2 is NOT included as part of the solution.

12. anonymous

So the solution is$\left\{ x|-3 \lt x \lt \frac{5}{2} \right\}$

13. anonymous

oh ok. thanks. what about x^2-4x greater than or equal to 0? i would do (x-2)(x+2)=0? then do the same thing for 2, -2?

14. anonymous

Well, you should see that you can factor x^2-4x > 0 as$x(x-4)=0$to solve for the partition points. The partitions will occur at x=0 and x=4. You'll have three intervals to test for again.

15. anonymous

Note, the inequality sign will tell you whether to include partition points. Here, you're not because you're dealing with a strict inequality. When it's not strict, check.

16. anonymous

so how do i check the points? could you show how?

17. anonymous

You just plug them in to the inequality to see if it gives you something that's true or not.

18. anonymous

ok so pluggin in 0 it would be -4=0 which is not true?

19. anonymous

Well, you only use the equality to find the points. ONCE you've found them, go back to the inequality; the whole reason behind doing this is to find x's that satisfy the inequality.

20. anonymous

Here, for x=0,$0(4-x)=0>0$That's NOT true - a number CANNOT be bigger than itself.

21. anonymous

So 0 is NOT included.

22. anonymous

so use a number beyond four? like 8? so 32=0 which is not true?

23. anonymous

Yep, choose a number beyond for to test for the interval, 4<x<infinity

24. anonymous

*four

25. anonymous

Go back to the inequality..! Not the equality.

26. anonymous

so id end up with 0<x>4

27. anonymous

$8(4-8)=-32$which is NOT greater than 0, so all x's above x=4 are NOT included.

28. anonymous

awesome helped a lot thanks

29. anonymous

You should have$(-\infty, 0) \cup (4, \infty)$as your interval.

30. anonymous

It means everything on the number line, except for everything in between and including, 0 and 4.

31. anonymous

One thing: your notation above 0<x>4 shouldn't be used. In mathematics, you just write x<0 or x>4

32. anonymous

The notation 0<x>4 has no meaning...which might get you in trouble when it comes to assessment.