How do you solve #(x-3)^2<=1# using a sign chart?

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Answer 1 · 2017-02-18T07:27:56

The answer is #x in [2,4]#

We need

#a^2-b^2=(a+b)(a-b)#

Let's rewrite the inequality

#(x-3)^2<=1#

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

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

#(x-4)(x-2)<=0#

Let #f(x)=(x-4)(x-2)#

Now, we build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##2##color(white)(aaaaa)##4##color(white)(aaaaaa)##+oo#

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

#color(white)(aaaa)##x-4##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 [2,4]#