How do you solve $x^{2} > 4$ $x^2>4$ using a sign chart?

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How do you solve $x^{2} > 4$ $x^2>4$ using a sign chart?

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Answer 1 · 2016-12-16T12:46:34

The answer is $x \in] - \infty, - 2 [\cup] 2, + \infty [$ $x in ] -oo,-2 [ uu ] 2,+oo [$

Explanation:

Let $f (x) = x^{2} - 4$ $f(x)=x^2-4$

Let's rewrite the equation as

$x^{2} - 4 = (x + 2) (x - 2) > 0$ $x^2-4=(x+2)(x-2)>0$

Now we can do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 2$ $-2$ $a a a a$ $color(white)(aaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 2$ $x+2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 2$ $x-2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

Therefore,

$f (x) > 0$ $f(x)>0$ when $x \in] - \infty, - 2 [\cup] 2, + \infty [$ $x in ] -oo,-2 [ uu ] 2,+oo [$

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How do you solve $x^{2} > 4$ $x^2>4$ using a sign chart?

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