How do you solve $0 = 4 x^{3} + 2 x - 1$ $0 = 4x^3 + 2x-1$ ?

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How do you solve $0 = 4 x^{3} + 2 x - 1$ $0 = 4x^3 + 2x-1$ ?

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Answer 1 · 2016-03-10T20:55:30

Check there are no rational roots, then use Cardano's method to find the Real and two Complex roots.

Explanation:

Let $f (x) = 4 x^{3} + 2 x - 1$ $f(x) = 4x^3+2x-1$

By the rational root theorem, any roots of $4 x^{3} + 2 x - 1 = 0$ $4x^3+2x-1 = 0$ will be expressible in the form $\frac{p}{q}$ $p/q$ for integers $p$ $p$ and $q$ $q$ where $p$ $p$ is a divisor of the constant term $- 1$ $-1$ and $q$ $q$ a divisor of the coefficient $4$ $4$ of the leading term.

So the only possible rational roots are:

$\pm \frac{1}{4}$ $+-1/4$ , $\pm \frac{1}{2}$ $+-1/2$ , $\pm 1$ $+-1$

We find $f (x) \neq 0$ $f(x) != 0$ for all of these values, so $f (x) = 0$ $f(x) = 0$ has no rational roots.

Use Cardano's method, substituting $x = u + v$ $x = u + v$ to get:

$4 u^{3} + 4 v^{3} + 2 (6 u v + 1) (u + v) - 1 = 0$ $4u^3 + 4v^3 + 2(6uv+1)(u+v) - 1 = 0$

Add the constraint $v = - \frac{1}{6 u}$ $v = -1/(6u)$ to eliminate the term in $(u + v)$ $(u+v)$ to get:

$0 = 4 u^{3} - \frac{4}{{(6 u)}^{3}} - 1 = 4 u^{3} - \frac{1}{54 u^{3}} - 1$ $0 = 4u^3 - 4/(6u)^3 - 1 = 4u^3 - 1/(54u^3) - 1$

Multiply through by $54 u^{3}$ $54u^3$ to get:

$216 {(u^{3})}^{2} - 54 (u^{3}) - 1 = 0$ $216(u^3)^2-54(u^3)-1 = 0$

Use the quadratic formula to find:

$u^{3} = \frac{54 \pm \sqrt{54^{2} + 4 \cdot 216}}{2 \cdot 216}$ $u^3 = (54+-sqrt(54^2+4*216))/(2*216)$

$= \frac{54 \pm \sqrt{3780}}{432}$ $=(54+-sqrt(3780))/432$

$= \frac{54 \pm 6 \sqrt{105}}{432}$ $=(54+-6sqrt(105))/432$

$= \frac{27 \pm 3 \sqrt{105}}{216}$ $=(27+-3sqrt(105))/216$

Due to the symmetry of the derivation in $u$ $u$ and $v$ $v$ , we can take one of these roots as $u^{3}$ $u^3$ and the other as $v^{3}$ $v^3$ to find the Real root:

$x_{1} = \frac{1}{6} (\sqrt[3]{27 + 3 \sqrt{105}} + \sqrt[3]{27 - 3 \sqrt{105}})$ $x_1 = 1/6(root(3)(27+3sqrt(105))+root(3)(27-3sqrt(105)))$

and Complex roots:

$x_{2} = \frac{1}{6} (ω \sqrt[3]{27 + 3 \sqrt{105}} + ω^{2} \sqrt[3]{27 - 3 \sqrt{105}})$ $x_2 = 1/6(omega root(3)(27+3sqrt(105))+omega^2 root(3)(27-3sqrt(105)))$

$x_{3} = \frac{1}{6} (ω^{2} \sqrt[3]{27 + 3 \sqrt{105}} + ω \sqrt[3]{27 - 3 \sqrt{105}})$ $x_3 = 1/6(omega^2 root(3)(27+3sqrt(105))+omega root(3)(27-3sqrt(105)))$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ $1$ .

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How do you solve $0 = 4 x^{3} + 2 x - 1$ $0 = 4x^3 + 2x-1$ ?

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