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Question Number 114806 by Algoritm last updated on 21/Sep/20

Answered by MJS_new last updated on 21/Sep/20

$$a-8=\frac{\sqrt{24}}{\sqrt{a}}$$$$\mathrm{let}a=t^2\wedge t>0$$$$t^2-8=\frac{2\sqrt{6}}{t}$$$$t^3-8t-2\sqrt{6}=0$$$$\mathrm{as}\mathrm{always},\mathrm{try}\mathrm{factors}\mathrm{of}\mathrm{the}\mathrm{constant}$$$$\Rightarrow t_1=-\sqrt{6}$$$$(t+\sqrt{6})(t^2-\sqrt{6}t-2)=0$$$$\Rightarrow t_{2,3}=\frac{\sqrt{6}}{2}\pm \frac{\sqrt{14}}{2}$$$$\mathrm{but}t>0\Rightarrow t=\frac{\sqrt{6}}{2}+\frac{\sqrt{14}}{2}$$$$\Rightarrow a=5+\sqrt{21}$$$$\Rightarrow a-\sqrt{6a}=a-\sqrt{6}t=2$$

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