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

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

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Answer 1 · 2016-12-21T19:40:31

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

Explanation:

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

Now we can do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

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

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

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

Therefore,

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

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

1 Answer

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How do you solve 9−x2<09-x^2<0 using a sign chart?

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Explanation:

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