How do you solve #7-x^2<=0#?

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Answer 1 · 2017-02-15T09:02:48

The answer is #x in ]-oo,-sqrt7]uu[sqrt7,+oo[#

Let's factorise the inequality

#7-x^2<=0#

#(sqrt7+x)(sqrt7-x)<=0#

Let #f(x)=(sqrt7+x)(sqrt7-x)#

Now, we build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-sqrt7##color(white)(aaaa)##sqrt7##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##sqrt7+x##color(white)(aaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##sqrt7-x##color(white)(aaaaaa)##+##color(white)(aaaaa)##+##color(white)(aaaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaa)##-#

Therefore,

#f(x)<=0# when #x in ]-oo,-sqrt7]uu[sqrt7,+oo[#