Question

Solve for x: Formula1" title=" 2x^{2/3} + 3x^{1/3} - 2 = 0 " alt=" 2x^{2/3} + 3x^{1/3} - 2 = 0 " align="absmiddle" class="latex-formula">

Answer 1

lemme post just to add to the superb reply above by @Lammetthash

$\bf 2x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0\implies \stackrel{\textit{notice is just }ax^2+bx+c=0}{2\left( x^{\frac{1}{3}} \right)^2+3\left( x^{\frac{1}{3}} \right)-2}=0\\\\\\\left(2x^{\frac{1}{3}}-1 \right)\left(x^{\frac{1}{3}}+2 \right)=0\\\\-------------------------------$

$\bf 2x^{\frac{1}{3}}-1=0\implies 2x^{\frac{1}{3}}=1\implies x^{\frac{1}{3}}=\cfrac{1}{2}\implies x=\left( \cfrac{1}{2} \right)^3\\\\\\x=\cfrac{1^3}{2^3}\implies \boxed{x=\cfrac{1}{8}}\\\\-------------------------------\\\\x^{\frac{1}{3}}+2=0\implies x^{\frac{1}{3}}=-2\implies x=(-2)^3\implies \boxed{x=-8}$

Answer 2

$2x^{2/3}+3x^{1/3}-2=(2x^{1/3}-1)(x^{1/3}+2)=0$

Then

$2x^{1/3}-1=0\implies2x^{1/3}=1\implies x^{1/3}=\dfrac12\implies x=\dfrac18$

or

$x^{1/3}+2=0\implies x^{1/3}=-2\implies x=-8$

(assuming you are solving over the real numbers)