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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question