Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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)

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com