How do you solve #1/2x^2<=x+12# using a sign chart?

1 Answer
Aug 9, 2017

Answer:

The solution is #x in [-4 ,6]#

Explanation:

Let's rearrange the inequality

#1/2x^2<= x+12#

#1/2x^2- x-12<=0#

#x^2- 2x-24 <=0#

Factorising, we get

#(x+4)(x-6) <=0#

Let, #f(x)=(x+4)(x-6)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##+oo##color(white)(aaaaa)##-4##color(white)(aaaaaa)##6##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aa)##0##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-6##color(white)(aaaaaa)##-##color(white)(aa)##color(white)(aaaa)##-##color(white)(a)##0##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aa)##0##color(white)(aaa)##-##color(white)(a)##0##color(white)(aa)##+#

Therefore,

#f(x) <=0# when #x in [-4 ,6]#

graph{1/2x^2-x-12 [-16.22, 15.8, -12.55, 3.47]}