How do you solve #5x<=2-3x^2# using a sign chart?

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Answer 1 · 2017-04-23T06:47:26

The solution is #x in [-2,1/3]#

Let's rearrange and factorise the inequality

#5x<=2-3x^2#

#3x^2+5x-2<=0#

#(3x-1)(x+2)<=0#

Let #f(x)=(3x-1)(x+2)#

We build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaa)##1/3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##3x-1##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x9##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in [-2,1/3]#