Question Number 209986 by a.lgnaoui last updated on 28/Jul/24
$$\mathrm{Solve}\: \\ $$$$\:\boldsymbol{\mathrm{ax}}^{\mathrm{3}} −\boldsymbol{\mathrm{bx}}\sqrt{\boldsymbol{\mathrm{x}}}\:+\boldsymbol{\mathrm{c}}=\mathrm{0}\:\:\:\:\: \\ $$$$\:\left(\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}}\right)\in\mathbb{R}^{\mathrm{3}} \:\:\:\:\mathrm{and}\:\boldsymbol{\mathrm{x}}\in\mathbb{R} \\ $$$$\left(\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{x}}\:\boldsymbol{{for}}\:\boldsymbol{{a}}=\mathrm{1},\:\:\boldsymbol{{b}}=\mathrm{9},\boldsymbol{{c}}=\mathrm{8}\right) \\ $$
Answered by mr W last updated on 28/Jul/24
$${let}\:{t}={x}\sqrt{{x}} \\ $$$${at}^{\mathrm{2}} −{bt}+{c}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\Rightarrow{x}={t}^{\frac{\mathrm{2}}{\mathrm{3}}} =\left(\frac{{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$
Commented by a.lgnaoui last updated on 28/Jul/24
$$\mathrm{thanks} \\ $$