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 $x>1+sqrt2$

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