anonymous
  • anonymous
how do you solve (2x^2-7x+3)/(6x^2+x-2)<0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Hero
  • Hero
First factor \(2x^2 - 7x + 3\) and \(6x^2 + x - 2\)
anonymous
  • anonymous
then what do i do?
Hero
  • Hero
First factor each trinomial. Let me know what you get for each.

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anonymous
  • anonymous
\[(2x-1)(x-3)/(3x+2)(2x-1)\]
Hero
  • Hero
Next let 1/2, 3, and -3/2 be your critical points.
Hero
  • Hero
Create a number line with those critical points on it. |dw:1438828606993:dw|
anonymous
  • anonymous
ok i understand all of that but how am i supposed to answer this?
Hero
  • Hero
Now you have to find the proper intervals that make \[\dfrac{(2x - 1)(x - 3)}{(2x - 1)(3x + 2)} < 0\] true
anonymous
  • anonymous
sot th only answer is -3/2 ?
Hero
  • Hero
We haven't gotten to "answers" yet. Hold down ctrl and press the minus key to zoom out.
anonymous
  • anonymous
ok
Hero
  • Hero
If you zoom out, you'll see that I created a number line with -3/2, 0, 1/2, and 3 on it.
anonymous
  • anonymous
yep i see that
Hero
  • Hero
So, basically, the next step is to pick a number between any of those critical points and "test" it.
anonymous
  • anonymous
so btwn -3/2 and 0, i just plug in -1?
Hero
  • Hero
To see if the expression on the left of the inequality is less than zero.
anonymous
  • anonymous
ok if i plug -1 in it positive
anonymous
  • anonymous
ok if i plug -1 in it positive
Hero
  • Hero
You should start with x = 0. Zero is not a critical point so you can test it.
anonymous
  • anonymous
i geta negative number
Hero
  • Hero
And that negative number is obviously less than zero, which means the interval (-3/2, 1/2) is one of the solutions.
anonymous
  • anonymous
so do i write the answer like -3/2
Hero
  • Hero
Now understand that the solution will never include two intervals back to back so, try any number greater than 3 to check if 3 < x < ∞ is the other interval.
anonymous
  • anonymous
no its positive
Hero
  • Hero
Which number did you test?
anonymous
  • anonymous
4
Hero
  • Hero
Test the number 2 and see what you get.
anonymous
  • anonymous
i get a negative
Hero
  • Hero
Basically, testing 2 confirms that 1/2 is not a critical point because 2x - 1 is a factor of one.
Hero
  • Hero
So the proper solution interval is actually -3/2 < x < 3
Hero
  • Hero
If you have factors of one in your factorizations, just cancel them out and don't include factors of one in your testing sequence. What I mean by that is, we had \[\dfrac{(2x - 1)(x - 3)}{(2x - 1)(3x + 2)} < 0\] And we should simplify that properly by cancelling out 2x - 1 to get \[\dfrac{x - 3}{3x + 2} < 0\]
Hero
  • Hero
So that's the expression we should have used to test the intervals. And we would only need to use -3/2 and 3 as our critical points.
anonymous
  • anonymous
ok i get it! so my final answer would be -3/2
Hero
  • Hero
The simplified expression (x - 3)/(3x + 2) for the given inequality is less than zero on the interval -3/2 < x < 3. There are no equations here.
anonymous
  • anonymous
oh ok ! thanks so much for explaining all of this to me!!!!! you rock!
Hero
  • Hero
You're most welcome.

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