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]