How do you solve #x^2-6x+9>16# using a sign chart?

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Answer 1 · 2017-03-01T20:37:22

The solution is #x in ]-oo,-1[uu]7,+oo[#

Let's rewrite the inequality and factorise

#x^2-6x+9-16>0#

#x^2-6x-7>0#

#(x-7)(x+1)>0#

Let #f(x)=(x-7)(x+1)#

We build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##7##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x-7##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,-1[uu]7,+oo[#