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.

How do I use a sign chart to solve 4x^2<=28x4x^2<=28x?