How do you solve #16p^2-81p^6<=0# using a sign chart?

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Answer 1 · 2017-01-27T11:06:21

The answer is #p in ]-oo, -2/3] uu [2/3, +oo [#

We need

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

Let's factorise the inequality

#16p^2-81p^6=p^2(16-81p^4)#

#=p^2(4-9p^2)(4+9p^2)#

#=p^2(2+3p)(2-3p)(4+9p^2)#

Let #f(p)=p^2(2+3p)(2-3p)(4+9p^2)#

#p^2>=0 , AA p in RR#

#4+9p^2>0 , AA p in RR#

Now we can build the sign chart

#color(white)(aaaa)##p##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-2/3##color(white)(aaaaaa)##2/3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##2+3p##color(white)(aaaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##2-3p##color(white)(aaaaaaa)##+##color(white)(aaaaa)##+##color(white)(aaaaa)##-#

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

Therefore,

#f(p)<=0# when #p in ]-oo, -2/3] uu [2/3, +oo [#