How do I use a sign chart to solve #2x^2-7x> -3#?

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Answer 1 · 2018-04-04T11:46:41

Solution: # x <0.5 and x>3 or (-oo,0.5)uu(3,oo) #

#2x^2-7x> -3 or 2x^2-7x +3 > 0 # or

#2x^2-6x-x +3 > 0 # or

#2x(x-3)-1(x -3) > 0 # or

#(x-3)(2x-1) > 0#

Critical points:

#(x-3)(2x-1)=0# . Critical points are #2x-1=0:. x= 1/2 #

and #x-3=0:. x= 3:.# critical points are #1/2,3#

Sign chart:

When #x< 1/2# sign of #(x-3)(2x-1) # is # (-) * (-) = (+) ; > 0#

When # 1/2 < x < 3 # sign of #(x-3)(2x-1) # is # (-) * (+) = (-) ; < 0#

When #x> 3# sign of #(x-3)(2x-1) # is # (+) * (+) = (+) ; > 0#

Solution: # x <0.5 and x>3 or (-oo,0.5)uu(3,oo) # [Ans]