How do you solve $x^{2} \leq 8 x$ $x^2<=8x$ using a sign chart?

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How do you solve $x^{2} \leq 8 x$ $x^2<=8x$ using a sign chart?

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Answer 1 · 2016-12-06T09:50:13

The answer is $x \in [0, 8]$ $x in [0,8]$

Explanation:

Let's rearrange the equation

$x^{2} \leq 8 x$ $x^2<=8x$

$x^{2} - 8 x \leq 0$ $x^2-8x<=0$

$x (x - 8) \leq 0$ $x(x-8)<=0$

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

Let's do the sign chart

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

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a a$ $color(white)(aaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 8$ $x-8$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

So, $f (x) \leq 0$ $f(x)<=0$ , when $x \in [0, 8]$ $x in [0,8]$

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

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How do you solve $x^{2} \leq 8 x$ $x^2<=8x$ using a sign chart?

1 Answer

Explanation:

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