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]}