How do you solve #x^2<=8x# using a sign chart?

1 Answer
Dec 6, 2016

The answer is #x in [0,8]#

Explanation:

Let's rearrange the equation

#x^2<=8x#

#x^2-8x<=0#

#x(x-8)<=0#

Let #f(x)=x(x-8)#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaa)##8##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x-8##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

So, #f(x)<=0#, when # x in [0,8]#

graph{x(x-8) [-41.1, 41.1, -20.56, 20.56]}