How do you solve x^3 − 2x > 1 ?

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Explanation:

First subtract 1 from both sides to get $x^{3} - 2 x - 1 > 0$ .

Notice that $x = - 1$ is a solution of $x^{3} - 2 x - 1 = 0$ , so $(x + 1)$ is a factor of $x^{3} - 2 x - 1$ . The other factor is $x^{2} - x - 1$ .

$x^{3} - 2 x - 1 = (x^{2} - x - 1) (x + 1)$ .

Solve $x^{2} - x - 1 = 0$ using the quadratic formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 1)}}{2 \cdot 1}$

$= \frac{1 \pm \sqrt{5}}{2}$

So $x^{3} - 2 x - 1$ will cross the $x$ -axis at $x = - 1$ , $x = \frac{1 - \sqrt{5}}{2}$ and $x = \frac{1 + \sqrt{5}}{2}$ .

Since the dominant term is $x^{3}$ , the function $x^{3} - 2 x - 1$ will be negative for $x < - 1$ , positive for $- 1 < x < \frac{1 - \sqrt{5}}{2}$ , negative for $\frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2}$ and positive for $x > \frac{1 + \sqrt{5}}{2}$ .

So the solution to the original problem is:

$- 1 < x < \frac{1 - \sqrt{5}}{2}$ or $x > \frac{1 + \sqrt{5}}{2}$ .

How do you solve x^3 − 2x > 1 ?

How do you solve x^3 − 2x > 1 ?

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