How do you solve #16x^2-81>=0#?

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Answer 1 · 2017-03-01T15:08:42

The solution is #x in ]-oo,-9/4]uu[9/4,+oo[#

We need

#a^2-b^2=(a+b)(a-b)#

Let's factorise the inequality

#16x^2-81=(4x+9)(4x-9)#

Let #f(x)=(4x+9)(4x-9)#

We build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-9/4##color(white)(aaaa)##9/4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##4x+9##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##4x-9##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)>=0# when #x in ]-oo,-9/4]uu[9/4,+oo[#