How do you solve #4x^4+4<=17x^2# using a sign chart?

1 Answer
Narad T.
Oct 15, 2016

#4x^4+4<=17x^2#
#4x^4-17x^2+4<=0#
#(4x^2-1)(x^2-4)<=0#
#(2x-1)(2x+1)(x+2)(x-2<=0#
# x=1/2 ; x=-1/2; x=-2;x=2#
Here is the sign chart
x -oo -2 -1/2 1/2 2 +oo
2x-1 - 0 + + +
2x+1 - - - 0 + +
X+2 - 0 + + + +
x-2 - - - - 0 +
(4x^4-17x^2+4) + - + - + +
#-2<=x<=-1/2#
and
#1/2<=x<=2#

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