How do you solve $x^3<=4x^2+3x$ using a sign chart?

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How do you solve $x^3<=4x^2+3x$ using a sign chart?

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Answer 1 · 2017-02-09T07:44:26

The answer is $x in ]-oo,2-sqrt7]uu [0,2+sqrt7]$

Explanation:

Let's rewrite and factorise the inequality

$x^3-4x^2-3x<=0$

$x(x^2-4x-3))<=0$

We calculate the roots of the equation $x^2-4x-3=0$

$Delta=(-4)^2-4*1*(-3)=16+12=28$

$x_1=(4+sqrt28)/2=(4+2sqrt7)/2=2+sqrt7$

$x_2=(4-sqrt28)/2=(4-2sqrt7)/2=2-sqrt7$

Let $f(x)=x(x-(2+sqrt7))(x-(2-sqrt7))$

Now, we build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaa)$ $-oo$ $color(white)(aaaa)$ $2-sqrt7$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $2+sqrt7$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-x_2$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaa)$ $+$ $color(white)(aaaaaaa)$ $+$

$color(white)(aaaa)$ $x-x_1$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaaaaa)$ $+$

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

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaaa)$ $-$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaaaaa)$ $+$

Therefore,

$f(x)<=0$ when $x in ]-oo,2-sqrt7]uu [0,2+sqrt7]$

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How do you solve $x^3<=4x^2+3x$ using a sign chart?

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