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

1 Answer
Mar 1, 2017

Answer:

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

Explanation:

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[#