How do you solve $x^3>2x^2+x$ using a sign chart?

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Answer 1 · 2017-08-25T12:46:26

Solution : $-0.4142 < x <0 **and**$ x >2.4142 $**In interval notation:**$ (-0.4142,0) uu (2.4142,oo)#

$x^3 > 2x^2 +x or x^3 - 2x^2 - x > 0 or x( x^2-2x-1) >0$ or

Roots of $(x^2-2x-1)$ are $x = (2 +- sqrt((2^2-4*1*-1)))/2$

or $x = 1+- sqrt2 or x = 2.4142 , x= - 0.4142$

$:. x( x^2-2x-1) >0 or x( x-2.4142)(x+0.4142) >0$

Critical points are $x=0 , x=2.4142 , x= -0.4142$

Sign chart:

When $x < -0.4142$ sign of $x( x-2.4142)(x+0.4142)$ is

$(-) * (-) * (-) = (-) ; <0$

When $-0.4142 < x <0$ sign of $x( x-2.4142)(x+0.4142)$ is

$(-) * (-) * (+) = (+) ; >0$

When $0 < x <2.4142$ sign of $x( x-2.4142)(x+0.4142)$ is

$(+) * (-) * (+) = (-) ; <0$

When $x >2.4142$ sign of $x( x-2.4142)(x+0.4142)$ is

$(+) * (+) * (+) = (+) ; >0$

Solution: $-0.4142 < x <0 and$ x >2.4142 #

In interval notation: $(-0.4142,0) uu (2.4142,oo)$

graph{x^3-2x^2-x [-10, 10, -5, 5]} [Ans]

How do you solve x^3>2x^2+xx^3>2x^2+x using a sign chart?