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thank you very much

 

https://prnt.sc/p1py1k

 Sep 4, 2019
 #1
avatar+111387 
+1

6x^2 - x < 2       subtract  2 from both sides

 

6x^2 - x  - 2  <  0           factor the left side

 

(2x  + 1) (3x -  2)  <  0        (1)

 

Set both  of the linear  factors  to   0  and solve for x

 

2x + 1  = 0              3x - 2   = 0

2x  = -1                   3x  = 2

x = -1/2                     x  = 2/3

 

We  have three possible intervals to test

 

(-inf, -1/2)   ( -1/2, 2/3)     (2/3, inf)

 

Either the first and third intervals solve (1)   or   the middle interval does

 

Pick  an   x value   on  ( -1/2, 2/3)   [ I'll choose x = 0 ]  and  note that

 

(2(0) + 1) ( 3(0) -2)  <  2     ????

 

(1) ( -2)  =   <  2     ???

 

-2  < 2      is true

 

So.....the interval   (-1/2, 2/3)   is the solution....

 

 

 

 

 

 

cool cool cool

 Sep 4, 2019
 #2
avatar+1148 
0

Another way of doing it...

You can find the x-intercepts by setting the equation to \(6x^2-x-2=0\), and \(x=-\frac{1}{2}, \frac{2}{3}\).

Since the equation is saying that \(6x^2-x-2<0,\) all values below the x-axis are answers, and therefore, the answer is

\((-\frac{1}{2}, \frac{2}{3})\).

 

You are very welcome!

:P

 Sep 4, 2019

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