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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0x=5/2 and x=3? just plot them?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, take a number line and plot those points. You'll see that it's partitioned into three regions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now pick a point in the first region that's easy to do the math with. So, maybe pick x=5.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And test in the original inequality.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right. i ended up shading the region between the two points becasue it includes all of them?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At 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
 5 years ago
Best ResponseYou've already chosen the best response.0No...don't shade...it's just a visual to assist.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The next interval contains 0, so use that. Then the lefthand side is 0, which IS less than 15, so this interval contains x's that solve the inequality.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0So the solution is\[\left\{ x3 \lt x \lt \frac{5}{2} \right\}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok. thanks. what about x^24x greater than or equal to 0? i would do (x2)(x+2)=0? then do the same thing for 2, 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you should see that you can factor x^24x > 0 as\[x(x4)=0\]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
 5 years ago
Best ResponseYou've already chosen the best response.0Note, 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
 5 years ago
Best ResponseYou've already chosen the best response.0so how do i check the points? could you show how?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just plug them in to the inequality to see if it gives you something that's true or not.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so pluggin in 0 it would be 4=0 which is not true?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Here, for x=0,\[0(4x)=0>0\]That's NOT true  a number CANNOT be bigger than itself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 0 is NOT included.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so use a number beyond four? like 8? so 32=0 which is not true?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep, choose a number beyond for to test for the interval, 4<x<infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Go back to the inequality..! Not the equality.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so id end up with 0<x>4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[8(48)=32\]which is NOT greater than 0, so all x's above x=4 are NOT included.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome helped a lot thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should have\[(\infty, 0) \cup (4, \infty)\]as your interval.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It means everything on the number line, except for everything in between and including, 0 and 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One thing: your notation above 0<x>4 shouldn't be used. In mathematics, you just write x<0 or x>4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The notation 0<x>4 has no meaning...which might get you in trouble when it comes to assessment.
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