# How do you solve x^(2/3) - 3x^(1/3) - 4 = 0?

Jan 14, 2016

Set $z = {x}^{\frac{1}{3}}$ When you find the $z$ roots, find $x = {z}^{3}$

Roots are $\frac{729}{8}$ and $- \frac{1}{8}$

#### Explanation:

Set ${x}^{\frac{1}{3}} = z$

${x}^{\frac{2}{3}} = {x}^{\frac{1}{3} \cdot 2} = {\left({x}^{\frac{1}{3}}\right)}^{2} = {z}^{2}$

So the equation becomes:

${z}^{2} - 3 z - 4 = 0$

Δ=b^2-4ac

Δ=(-3)^2-4*1*(-4)

Δ=25

z_(1,2)=(-b+-sqrt(Δ))/(2a)

${z}_{1 , 2} = \frac{- \left(- 4\right) \pm \sqrt{25}}{2 \cdot 1}$

${z}_{1 , 2} = \frac{4 \pm 5}{2}$

${z}_{1} = \frac{9}{2}$

${z}_{2} = - \frac{1}{2}$

To solve for $x$:

${x}^{\frac{1}{3}} = z$

${\left({x}^{\frac{1}{3}}\right)}^{3} = {z}^{3}$

$x = {z}^{3}$

${x}_{1} = {\left(\frac{9}{2}\right)}^{3}$

${x}_{1} = \frac{729}{8}$

${x}_{2} = {\left(- \frac{1}{2}\right)}^{3}$

${x}_{2} = - \frac{1}{8}$

Jan 14, 2016

x = 64 or x = -1

#### Explanation:

note that ${\left({x}^{\frac{1}{3}}\right)}^{2} = {x}^{\frac{2}{3}}$

Factorising ${x}^{\frac{2}{3}} - 3 {x}^{\frac{1}{3}} - 4 = 0$ gives ;

$\left({x}^{\frac{1}{3}} - 4\right) \left({x}^{1 / 3} + 1\right) = 0$

$\Rightarrow \left({x}^{\frac{1}{3}} - 4\right) = 0 \mathmr{and} \left({x}^{\frac{1}{3}} + 1\right) = 0$

$\Rightarrow {x}^{\frac{1}{3}} = 4 \mathmr{and} {x}^{\frac{1}{3}} = - 1$

'cubing' both sides of the pair of equations :

${\left({x}^{\frac{1}{3}}\right)}^{3} = {4}^{3} \mathmr{and} {\left({x}^{\frac{1}{3}}\right)}^{3} = {\left(- 1\right)}^{3}$

$\Rightarrow x = 64 \mathmr{and} x = - 1$