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

Question

1359 views around the world

You can reuse this answer
Creative Commons License

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 $x in [2,4]$

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