How do you solve $4 x^{2} > 12 x - 9$ $4x^2>12x-9$ ?

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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,

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

How do you solve 4x^2>12x-94x2>12x−94x^2>12x-9?