# How do you find the solution to the quadratic equation 2x^(2/3) + 5^(1/3) = 12?

May 9, 2015

In this way:

let ${x}^{\frac{1}{3}} = t$ so ${x}^{\frac{2}{3}} = {t}^{2}$.

$2 {t}^{2} + 5 t - 12 = 0$

$\Delta = {b}^{2} - 4 a c = 25 + 4 \cdot 2 \cdot 12 = 121$

$t = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 5 \pm 11}{4} \Rightarrow$

${t}_{1} = \frac{- 5 - 11}{4} = \frac{- 16}{4} = - 4$

${t}_{2} = \frac{- 5 + 11}{4} = \frac{6}{4} = \frac{3}{2}$.

And now:

${x}^{\frac{1}{3}} = - 4 \Rightarrow {\left({x}^{\frac{1}{3}}\right)}^{3} = {\left(- 4\right)}^{3} \Rightarrow x = - 64$

${x}^{\frac{1}{3}} = \frac{3}{2} \Rightarrow {\left({x}^{\frac{1}{3}}\right)}^{3} = {\left(\frac{3}{2}\right)}^{3} \Rightarrow x = \frac{27}{8}$.