How do you solve and graph $x^{3} > 2 x^{2} + x$ $x^3>2x^2+x$ ?

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How do you solve and graph $x^{3} > 2 x^{2} + x$ $x^3>2x^2+x$ ?

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Answer 1 · 2018-05-31T04:11:36

The answers are: $1 - \sqrt{2} < x < 0$ $1-sqrt2 < x <0$ and $x > 1 + \sqrt{2}$ $x>1+sqrt2$

Explanation:

$x^{3} > 2 x^{2} + x$ $x^3>2x^2+x$

$x^{3} - 2 x^{2} - x > 0$ $x^3-2x^2-x>0$

$x (x^{2} - 2 x - 1) > 0$ $x(x^2-2x-1)>0$

Solve $x^{2} - 2 x - 1$ $x^2-2x-1$ using the quadratic formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{2 \pm \sqrt{4 - 4 (1) (- 1)}}{2}$ $x = (2+-sqrt(4-4(1)(-1)))/(2)$

$x = \frac{2 \pm \sqrt{8}}{2}$ $x = (2+-sqrt8)/2$

$x = \frac{2 \pm 2 \sqrt{2}}{2}$ $x=(2+-2sqrt2)/2$

$x = 1 \pm \sqrt{2}$ $x=1+-sqrt2$

So:
$x (x^{2} - 2 x - 1) > 0$ $x(x^2-2x-1)>0$

$x (x - (1 + \sqrt{2})) (x - (1 - \sqrt{2})) = 0$ $x(x-(1+sqrt2))(x-(1-sqrt2))=0$ is the graph below

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

$x (x - (1 + \sqrt{2})) (x - (1 - \sqrt{2})) > 0$ $x(x-(1+sqrt2))(x-(1-sqrt2))>0$ means that you need to find the parts of the graph that is above the line $y = 0$ $y=0$

Therefore, the answers are: $1 - \sqrt{2} < x < 0$ $1-sqrt2 < x <0$ and $x > 1 + \sqrt{2}$ $x>1+sqrt2$

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How do you solve and graph $x^{3} > 2 x^{2} + x$ $x^3>2x^2+x$ ?

1 Answer

Explanation:

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How do you solve and graph $x^{3} > 2 x^{2} + x$ $x^3>2x^2+x$ ?