How do you solve #4x^2>12x-9#?

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Answer 1 · 2017-02-15T09:23:45

The answer is # x in RR-{3/2}#

Let's rearrange and factorise the inequality

#4x^2>12x+9#

#4x^2-12x+9>0#

#(2x-3)(2x-3)>0#

#(2x-3)^2>0#

Let #f(x)=(2x-3)^2#

Therefore,

#f(x)>0 ; AA x in RR-{3/2}#