Question Number 22244 by ajfour last updated on 14/Oct/17
$${Solve}\:{the}\:{inequality}\:: \\ $$$$\:−\mathrm{9}\left(\sqrt[{\mathrm{4}}]{{x}}\right)+\sqrt{{x}}+\mathrm{18}\:\geqslant\:\mathrm{0}\:. \\ $$
Answered by $@ty@m last updated on 14/Oct/17
![Let (√x)=y −9(√y)+y+18≥0 9(√y)≤18+y 81y≤324+y^2 +36y y^2 −45y+324≥0 (y−9)(y−36)≥0 y∈(−∞,9]∪[36,∞) x∈(0,81]∪[1296,∞)](Q22250.png)
$${Let}\:\sqrt{{x}}={y} \\ $$$$−\mathrm{9}\sqrt{{y}}+{y}+\mathrm{18}\geqslant\mathrm{0} \\ $$$$\mathrm{9}\sqrt{{y}}\leqslant\mathrm{18}+{y} \\ $$$$\mathrm{81}{y}\leqslant\mathrm{324}+{y}^{\mathrm{2}} +\mathrm{36}{y} \\ $$$${y}^{\mathrm{2}} −\mathrm{45}{y}+\mathrm{324}\geqslant\mathrm{0} \\ $$$$\left({y}−\mathrm{9}\right)\left({y}−\mathrm{36}\right)\geqslant\mathrm{0} \\ $$$${y}\in\left(−\infty,\mathrm{9}\right]\cup\left[\mathrm{36},\infty\right) \\ $$$${x}\in\left(\mathrm{0},\mathrm{81}\right]\cup\left[\mathrm{1296},\infty\right) \\ $$
Commented by ajfour last updated on 14/Oct/17
$${Thanks},\:{this}\:{matches}\:{with}\:{the} \\ $$$${answer}\:{given}. \\ $$