How do you solve $16-x^2>0$ using a sign chart?

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Answer 1 · 2017-05-11T06:00:15

The solution is $x in (-4,+4)$

We need

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

We factorise the inequality

$16-x^2>0$

$(4+x)(4-x)>0$

Let

$f(x)=(4+x)(4-x)$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaa)$ $+4$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $4+x$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $4-x$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

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

Therefore,

$f(x)>0$ when $x in (-4,+4)$

How do you solve 16-x^2>016-x^2>0 using a sign chart?