How do you solve and graph #x^3>2x^2+x#?

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Answer 1 · 2018-05-31T04:11:36

The answers are: #1-sqrt2 < x <0# and #x>1+sqrt2#

#x^3>2x^2+x#

#x^3-2x^2-x>0#

#x(x^2-2x-1)>0#

Solve #x^2-2x-1# using the quadratic formula

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (2+-sqrt(4-4(1)(-1)))/(2)#

#x = (2+-sqrt8)/2#

#x=(2+-2sqrt2)/2#

#x=1+-sqrt2#

So:
#x(x^2-2x-1)>0#

#x(x-(1+sqrt2))(x-(1-sqrt2))=0# is the graph below

graph{x(x-(1+sqrt2))(x-(1-sqrt2)) [-10, 10, -5, 5]}

#x(x-(1+sqrt2))(x-(1-sqrt2))>0# means that you need to find the parts of the graph that is above the line #y=0#

Therefore, the answers are: #1-sqrt2 < x <0# and #x>1+sqrt2#