How do I use a sign chart to solve #4x^2<=28x#?

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Answer 1 · 2014-10-21T10:56:55

By subtracting #28x#,

#4x^2 le 28x Rightarrow 4x^2-28x le 0#

Let #f(x)=4x^2-28x#.

First, turn it into an equation.

#Rightarrow 4x^2-28x=0#

by factoring out #4x#,

#Rightarrow 4x(x-7)=0 Rightarrow x=0,7#

Use #0# and #7# to split the real number line into three intervals

#(-infty,0)#, #(0,7)#, and #(7,infty)#.

Choose any number from each interval as a sample point.
I choose #x=-1,1,8#

#f(-1)=4(-1)^2-28(-1)=32>0#

#f(1)=4(1)^2-28(1)=-24<0#

#f(8)=4(8)^2-28(8)=32>0#

The above indicates that #f(x)<0# on #(0,7)#, which means that #f(x) le 0# on #[0,7]# since #f(0)=f(7)=0#.

Hence, the solution set of the inequality is #[0,7]#.

I hope that this was helpful.