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]