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Given-f-x-9-5x-and-f-a-4a-2-find-the-possible-value-of-a-




Question Number 136861 by bramlexs22 last updated on 27/Mar/21
Given f((√(x+9)) )= 5x and f(a)=4a^2   find the possible value of a.
$$\mathrm{Given}\:\mathrm{f}\left(\sqrt{\mathrm{x}+\mathrm{9}}\:\right)=\:\mathrm{5x}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{a}\right)=\mathrm{4a}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}. \\ $$$$ \\ $$
Answered by EDWIN88 last updated on 27/Mar/21
 we have  { ((a=(√(x+9)) , x=a^2 −9)),((4a^2  = 5x , 4a^2 =5a^2 −45)) :}    a^2  = 45 , since (√(x+9)) ≥ 0 for ∀x∈Domain f so  the possible value of a only (√(45)) = 3(√5)
$$\:\mathrm{we}\:\mathrm{have}\:\begin{cases}{\mathrm{a}=\sqrt{\mathrm{x}+\mathrm{9}}\:,\:\mathrm{x}=\mathrm{a}^{\mathrm{2}} −\mathrm{9}}\\{\mathrm{4a}^{\mathrm{2}} \:=\:\mathrm{5x}\:,\:\mathrm{4a}^{\mathrm{2}} =\mathrm{5a}^{\mathrm{2}} −\mathrm{45}}\end{cases} \\ $$$$ \:\mathrm{a}^{\mathrm{2}} \:=\:\mathrm{45}\:,\:\mathrm{since}\:\sqrt{\mathrm{x}+\mathrm{9}}\:\geqslant\:\mathrm{0}\:\mathrm{for}\:\forall\mathrm{x}\in\mathrm{Domain}\:\mathrm{f}\:\mathrm{so} \\ $$$$\mathrm{the}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{only}\:\sqrt{\mathrm{45}}\:=\:\mathrm{3}\sqrt{\mathrm{5}}\: \\ $$

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