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 $z in ] 1,5 [$

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