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What is the greatest integer value of x for which 6x^2 + x - 2 < 0?

 Mar 27, 2020
 #1
avatar+1970 
+1

We will use the quadratic formula:

 

(-b+-sqrt(b^2-4ac))/2a

 

(-1+-sqrt(1+48))/12

=(-1+-7)/12

 

x=1/2 

x=-2/3

 

Since 1/2 is greater, the answer will be 1/2.

 

Hope this helped!

 Mar 27, 2020
 #2
avatar
+1

 

What is the greatest integer value of x for which 6x^2 + x - 2 < 0?   

 

This will factor to ( 3x + 2 )( 2x – 1 )

 

So you get      3x = –2   therefore   x = –2/3 

and you get     2x = +1   therefore   x = +1/2 

.

 Mar 27, 2020

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