anonymous
  • anonymous
2x^2+x<15 polynomial inequality how do i solve and graph on a real number line?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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.
anonymous
  • anonymous
x=5/2 and x=3? just plot them?
anonymous
  • anonymous
x=5/2 and x=-3.

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anonymous
  • anonymous
Yes, take a number line and plot those points. You'll see that it's partitioned into three regions.
anonymous
  • anonymous
Now pick a point in the first region that's easy to do the math with. So, maybe pick x=-5.
anonymous
  • anonymous
And test in the original inequality.
anonymous
  • anonymous
right. i ended up shading the region between the two points becasue it includes all of them?
anonymous
  • 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).
anonymous
  • anonymous
No...don't shade...it's just a visual to assist.
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
So the solution is\[\left\{ x|-3 \lt x \lt \frac{5}{2} \right\}\]
anonymous
  • 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?
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
so how do i check the points? could you show how?
anonymous
  • anonymous
You just plug them in to the inequality to see if it gives you something that's true or not.
anonymous
  • anonymous
ok so pluggin in 0 it would be -4=0 which is not true?
anonymous
  • 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.
anonymous
  • anonymous
Here, for x=0,\[0(4-x)=0>0\]That's NOT true - a number CANNOT be bigger than itself.
anonymous
  • anonymous
So 0 is NOT included.
anonymous
  • anonymous
so use a number beyond four? like 8? so 32=0 which is not true?
anonymous
  • anonymous
Yep, choose a number beyond for to test for the interval, 4
anonymous
  • anonymous
*four
anonymous
  • anonymous
Go back to the inequality..! Not the equality.
anonymous
  • anonymous
so id end up with 04
anonymous
  • anonymous
\[8(4-8)=-32\]which is NOT greater than 0, so all x's above x=4 are NOT included.
anonymous
  • anonymous
awesome helped a lot thanks
anonymous
  • anonymous
You should have\[(-\infty, 0) \cup (4, \infty)\]as your interval.
anonymous
  • anonymous
It means everything on the number line, except for everything in between and including, 0 and 4.
anonymous
  • anonymous
One thing: your notation above 04 shouldn't be used. In mathematics, you just write x<0 or x>4
anonymous
  • anonymous
The notation 04 has no meaning...which might get you in trouble when it comes to assessment.

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