How do solve $x^2<=8x$ and write the answer as a inequality and interval notation?

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Answer 1 · 2016-12-16T07:30:58

The answer is $x in [0 ,8 ]$ or $0<=x<=8$

Let's rewrite the equation as

$x^2-8x<=0$ , $=>$ , $x(x-8)<=0$

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

Let's do a sign chart

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

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aaa)$ $+$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $x-8$ $color(white)(aaaaa)$ $-$ $color(white)(aaa)$ $-$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $+$ $color(white)(aaa)$ $-$ $color(white)(aaa)$ $+$

Therefore,

$f(x)<=0$ when $x in [0 ,8 ]$ , or $0<=x<=8$

How do solve x^2<=8xx^2<=8x and write the answer as a inequality and interval notation?