How do you solve #z^2-6z+7<2#?

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Answer 1 · 2017-01-18T09:03:37

The answer is #z in ] 1,5 [#

Let's rearrange the inequation

#z^2-6z+7<2#

#z^2-6z+5<0#

Let's factorise

#(z-5)(z-1)<0#

Let #f(z)=(z-5)(z-1)#

Now. we can do the sign chart

#color(white)(aaaa)##z##color(white)(aaaaa)##-oo##color(white)(aaaa)##1##color(white)(aaaaaa)##5##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##z-1##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##z-5##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(z)<0#, when #z in ] 1,5 [#