How do you solve #x^2<=4x# using a sign chart?

1 Answer
Nov 16, 2016

Answer:

Move every term to one side, factor the expression, and determine when each factor is#+#or#-#. Multiply these#+#and#-#intervals from the factors to get the#+#and#-#intervals for the product (and thus the original inequality).

Explanation:

#x^2<=4x#
#x^2-4x<=0# Move everything to one side
#x(x-4)<=0# Factor the expression

This is now written as the product of two factors, #x# and #x-4#. The values of #x# that make the factors #0# are the "points of interest", because it is only at these points where the value of #x^2-4x# could change from being#<=0# to not being#<=0#. We set each factor equal to #0# and solve for #x#:

#x=0 or x-4=0#
#=>x=0 or x=4#

So we have three intervals to examine: below 0, between 0 and 4, and above 4.

A sign chart can be made like this:
. . . . . . . . . .---------0----------4-----------
#x#
#x-4#
. . . . . . . . . .----------------------------------
#x(x-4)#

Fill in the sign chart with + and - signs to reflect where each factor is#<=0#. For example: the factor #x-4# is negative when #x<4#; it's 0 when #x=4#; and it's positive when #x>4#.

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

To complete the bottom row of the sign chart, each interval is filled with the product of all the signs above it. For example, when #x<0#, #(-)# x #(-)# = #(+)#.

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

This tells you that #x(x-4)# (in other words, #x^2-4x#) will be negative only when #x# is between 0 and 4, and it will be 0 at those endpoints, i.e. #x^2-4x<=0# when #0<=x<=4#. And since #x^2-4x<=0# is equivalent to our original inequality #x^2<=4x#, we are done.

The solution is #0<=x<=4#.
Or #x in [0,4]# if you prefer interval notation.

Go ahead and try some values for yourself: say #x=3#. We should expect the inequality to be true. Well, is it?

#x^2<=4x#
#<=>3^2<=4(3)# (when #x=3#)
#<=>9<=12#
Which is true!