How do you solve the inequality #|6x^2-3|<7x#?

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Answer 1 · 2016-05-01T12:55:27

#x in(1/3. 3/2)#

The given inequality is the combined inequality for the pair

#6x^2-3<7x and -(6x^2-3)<7x#.

So, #(2x-3)(3x+1)<0 and (2x+3)(3x-1)>0#. These are equivalent to.

#(x-3/2)(x+1/3)<0 and (x+3/2)(x-1/3)>0#.

Use that, if #a < b and (x-a)(x-b)<0#, then #x in(a, b)#

and.if #a < b and (x-a)(x-b)>0#, then x < a and x > b#..

Accordingly, # x in (-1/3, 3/2)# and x does not # in[-3/2, 1/3]#.
So, # x in# (intersection of these two intervals)#=(1/3, 3/2)#.

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