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Question Number 136861 by bramlexs22 last updated on 27/Mar/21

$$\mathrm{Given}\mathrm{f}\left(\sqrt{\mathrm{x}+9}\right)=5\mathrm{x}\mathrm{and}\mathrm{f}\left(\mathrm{a}\right)={4\mathrm{a}}^2$$$$\mathrm{find}\mathrm{the}\mathrm{possible}\mathrm{value}\mathrm{of}\mathrm{a}.$$$$$$

Answered by EDWIN88 last updated on 27/Mar/21

$$\mathrm{we}\mathrm{have}\{\begin{array}{l}\mathrm{a}=\sqrt{\mathrm{x}+9},\mathrm{x}={\mathrm{a}}^2-9\\ {4\mathrm{a}}^2=5\mathrm{x},{4\mathrm{a}}^2={5\mathrm{a}}^2-45\end{array}$$$${\mathrm{a}}^2=45,\mathrm{since}\sqrt{\mathrm{x}+9}⩾0\mathrm{for}\mathrm{\forall }\mathrm{x}\in \mathrm{Domain}\mathrm{f}\mathrm{so}$$$$\mathrm{the}\mathrm{possible}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{only}\sqrt{45}=3\sqrt{5}$$

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