How do you solve #(x+5)(x-2)>=0#?

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Answer 1 · 2016-12-20T15:40:20

The answer is #x in ] -oo,-5 ] uu [2, +oo[#

Let #f(x)=(x+5)(x-2)#

Now, we can do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

#f(x)>=0#, when #x in ] -oo,-5 ] uu [2, +oo[#

graph{(x+5)(x-2) [-28.86, 28.85, -14.43, 14.43]}