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18-7x-x-2-2x-9-18-7x-x-2-x-8-




Question Number 80917 by jagoll last updated on 08/Feb/20
((√(18−7x−x^2 ))/(2x+9)) ≥ ((√(18−7x−x^2 ))/(x+8))
$$\frac{\sqrt{\mathrm{18}−\mathrm{7}{x}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}+\mathrm{9}}\:\geqslant\:\frac{\sqrt{\mathrm{18}−\mathrm{7}{x}−{x}^{\mathrm{2}} }}{{x}+\mathrm{8}} \\ $$
Answered by john santu last updated on 08/Feb/20
(i) x^2 −7x−18≤0 ⇒−9≤x≤2  (ii) (1/(2x+9)) ≥ (1/(x+8))  ((x+1)/((2x+9)(x+8))) ≤0   x< −8∨−(9/2)<x≤−1  solution (i)∧(ii)  ∴ −9≤x<−8 ∨ −(9/2)<x≤−1 ∧ x=2
$$\left({i}\right)\:{x}^{\mathrm{2}} −\mathrm{7}{x}−\mathrm{18}\leqslant\mathrm{0}\:\Rightarrow−\mathrm{9}\leqslant{x}\leqslant\mathrm{2} \\ $$$$\left({ii}\right)\:\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{9}}\:\geqslant\:\frac{\mathrm{1}}{{x}+\mathrm{8}} \\ $$$$\frac{{x}+\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{9}\right)\left({x}+\mathrm{8}\right)}\:\leqslant\mathrm{0}\: \\ $$$${x}<\:−\mathrm{8}\vee−\frac{\mathrm{9}}{\mathrm{2}}<{x}\leqslant−\mathrm{1} \\ $$$${solution}\:\left({i}\right)\wedge\left({ii}\right) \\ $$$$\therefore\:−\mathrm{9}\leqslant{x}<−\mathrm{8}\:\vee\:−\frac{\mathrm{9}}{\mathrm{2}}<{x}\leqslant−\mathrm{1}\:\wedge\:{x}=\mathrm{2} \\ $$
Commented by jagoll last updated on 08/Feb/20
thx
$${thx} \\ $$

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