How do you solve $–x^2 + 4 < 0$ ?

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Answer 1 · 2018-07-01T11:40:19

The solution is $x in (-oo, -2) uu(2, +oo)$

The inequality is

$-x^2+4<0$

$=>$ , $x^2-4>0$

$=>$ , $(x+2)(x-2)>0$

Let $y=(x+2)(x-2)$

Build a sign chart

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

$color(white)(aaaa)$ $x+2$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $y$ $color(white)(aaaaaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$y>0$ when $x in (-oo, -2) uu(2, +oo)$

graph{-x^2+4 [-7.9, 7.9, -3.95, 3.95]}

How do you solve –x^2 + 4 < 0–x^2 + 4 < 0?