How do you solve #x^2+x-6>0#?

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Answer 1 · 2017-01-06T11:16:57

The answer is #x in ] -oo,-3[ uu ] 2, oo[ #

We start by factorising the expression

#x^2+x-6=(x-2)(x+3)#

and

let #f(x)=(x-2)(x+3)#

Now, we can make the sign chart

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

#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

#f(x)>0#, when # x in ] -oo,-3[ uu ] 2, oo[ #