How do you solve $9-x^2<0$ using a sign chart?

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Answer 1 · 2016-12-21T19:40:31

The answer is $x in ] -oo,-3 [ uu ] 3, oo[$

Let $f(x)=9-x^2=(3-x)(3+x)$

Now we can do the sign chart

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

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

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

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

Therefore,

$f(x)<0$ , when $x in ] -oo,-3 [ uu ] 3, oo[$

How do you solve 9-x^2<09-x^2<0 using a sign chart?