How do you solve $4x^2-9>0$ using a sign chart?

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Answer 1 · 2017-07-06T12:12:37

Solution : $x < -3/2 or x > 3/2$ , In interval notation $(-oo,-3/2)uu (3/2,oo)$

$4x^2-9 >0 or (2x)^2 - 3^2 >0 or$ (2x+3)(2x-3) >0 #

Critical points are $2x+3=0 or x= -3/2 and 2x-3=0 or x= 3/2$

Sign chart:

When $x < -3/2 ; (2x+3)(2x-3)$ sign is $(-)*(-) = + :. >0$

When $-3/2 < x < 3/2 ; (2x+3)(2x-3)$ sign is $(+)*(-) = - :. <0$

When $x > 3/2 ; (2x+3)(2x-3)$ sign is $(+)*(+) = + :. >0$

So Solution is $x < -3/2 or x > 3/2$ , In interval notation $(-oo,-3/2)uu (3/2,oo)$ [Ans]

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