Question

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

Answer 1

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.

Answer 2

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

Answer 3

x=5/2 and x=-3.

Answer 4

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

Answer 5

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

Answer 6

And test in the original inequality.

Answer 7

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

Answer 8

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).

Answer 9

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

Answer 10

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.

Answer 11

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.

Answer 12

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

Answer 13

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?

Answer 14

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.

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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.

Answer 20

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

Answer 21

So 0 is NOT included.

Answer 22

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

Answer 23

Yep, choose a number beyond for to test for the interval,

4<x<infinity

Answer 24

*four

Answer 25

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

Answer 26

so id end up with 0<x>4

Answer 27

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

Answer 28

awesome helped a lot thanks

Answer 29

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

Answer 30

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

Answer 31

One thing: your notation above

0<x>4

shouldn't be used. In mathematics, you just write

x<0 or x>4

Answer 32

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