How do you solve $x^2-4x<-3$ using a sign chart?

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Answer 1 · 2016-12-13T15:37:15

The answer is $x in ] 1,3 [$

Rewrite the equation as, $x^2-4x+3<0$

Let $f(x)=x^2-4x+3=(x-1)(x-3)$

The domain of $f(x)$ is $D_f(x)=RR$

Now, we can do our sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $+1$ $color(white)(aaaa)$ $+3$ $color(white)(aaaa)$ $+oo$

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

$color(white)(aaaa)$ $x-3$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

So,

$f(x)<0$ when $x in ] 1,3 [$

How do you solve x^2-4x<-3x^2-4x<-3 using a sign chart?