how do you solve
(2x^2-7x+3)/(6x^2+x-2)<0

- anonymous

how do you solve
(2x^2-7x+3)/(6x^2+x-2)<0

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- Hero

First factor \(2x^2 - 7x + 3\) and \(6x^2 + x - 2\)

- anonymous

then what do i do?

- Hero

First factor each trinomial. Let me know what you get for each.

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## More answers

- anonymous

\[(2x-1)(x-3)/(3x+2)(2x-1)\]

- Hero

Next let 1/2, 3, and -3/2 be your critical points.

- Hero

Create a number line with those critical points on it.
|dw:1438828606993:dw|

- anonymous

ok i understand all of that but how am i supposed to answer this?

- Hero

Now you have to find the proper intervals that make \[\dfrac{(2x - 1)(x - 3)}{(2x - 1)(3x + 2)} < 0\] true

- anonymous

sot th only answer is -3/2 ?

- Hero

We haven't gotten to "answers" yet. Hold down ctrl and press the minus key to zoom out.

- anonymous

ok

- 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

yep i see that

- Hero

So, basically, the next step is to pick a number between any of those critical points and "test" it.

- anonymous

so btwn -3/2 and 0, i just plug in -1?

- Hero

To see if the expression on the left of the inequality is less than zero.

- anonymous

ok if i plug -1 in it positive

- anonymous

ok if i plug -1 in it positive

- Hero

You should start with x = 0. Zero is not a critical point so you can test it.

- anonymous

i geta negative number

- Hero

And that negative number is obviously less than zero, which means the interval (-3/2, 1/2) is one of the solutions.

- anonymous

so do i write the answer like -3/2

- 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

no its positive

- Hero

Which number did you test?

- anonymous

4

- Hero

Test the number 2 and see what you get.

- anonymous

i get a negative

- Hero

Basically, testing 2 confirms that 1/2 is not a critical point because 2x - 1 is a factor of one.

- Hero

So the proper solution interval is actually -3/2 < x < 3

- 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

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

ok i get it! so my final answer would be -3/2

- 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

oh ok ! thanks so much for explaining all of this to me!!!!! you rock!

- Hero

You're most welcome.

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