How do you solve #1/2x^2>=4-x# using a sign chart?

Question

1148 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T11:30:03

Solution: # x<=-4 and x>=2 or (-oo, -4] uu [2,oo)#

#1/2 x^2 >= 4-x or x^2 >= 8- 2x or x^2 +2x -8 >=0 # or

# x^2 +4x-2x -8 >=0 or x(x+4) -2 (x+4) >= 0 or (x+4)(x-2) >=0#

Critical points are #x= -4 , x=2 #

Sign chart:
When # x < -4# # (x+4) (x-2) is (-)*(-) = (+) , >0#

When # -4 < x < 2 (x+4)(x-2) is (+)*(-)= - <0#

When # x > 2 # # (x+4) (x-2) is (+)*(+) = (+) >0#

Solution # x<=-4 and x>=2 or (-oo, -4] uu [2,oo)# [Ans]