How do you solve #16-x^2>0# using a sign chart?

1 Answer
May 11, 2017

Answer:

The solution is #x in (-4,+4)#

Explanation:

We need

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

We factorise the inequality

#16-x^2>0#

#(4+x)(4-x)>0#

Let

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

We can build the sign chart

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

#color(white)(aaaa)##4+x##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##4-x##color(white)(aaaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#

Therefore,

#f(x)>0# when #x in (-4,+4)#