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Question Number 43496 by LYCON TRIX last updated on 11/Sep/18

(√(a−(√(a+x))))  +  (√(a+(√(a−x))))  =2x   Solve for “ x ” in terms of  “  a  ”

$$\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}\:\:+\:\:\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}\:\:=\mathrm{2x}\: \\ $$$$\mathrm{Solve}\:\mathrm{for}\:``\:\mathrm{x}\:''\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\:``\:\:\mathrm{a}\:\:'' \\ $$$$ \\ $$

Commented by LYCON TRIX last updated on 11/Sep/18

NS7UC −standard question 58   I would like to know your   opinion and best possible solution on SQ−58   this question holds cash prize too !

$$\mathrm{NS7UC}\:−\mathrm{standard}\:\mathrm{question}\:\mathrm{58}\: \\ $$$$\mathrm{I}\:\mathrm{would}\:\mathrm{like}\:\mathrm{to}\:\mathrm{know}\:\mathrm{your}\: \\ $$$$\mathrm{opinion}\:\mathrm{and}\:\mathrm{best}\:\mathrm{possible}\:\mathrm{solution}\:\mathrm{on}\:\mathrm{SQ}−\mathrm{58}\: \\ $$$$\mathrm{this}\:\mathrm{question}\:\mathrm{holds}\:\mathrm{cash}\:\mathrm{prize}\:\mathrm{too}\:! \\ $$$$ \\ $$

Commented by LYCON TRIX last updated on 11/Sep/18

i′ll be happy to see whole solution

$$\mathrm{i}'\mathrm{ll}\:\mathrm{be}\:\mathrm{happy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{whole}\:\mathrm{solution} \\ $$

Commented by LYCON TRIX last updated on 11/Sep/18

its incomplete or invalid   Q Code − 42408 , 42471 and 43496

$$\mathrm{its}\:\mathrm{incomplete}\:\mathrm{or}\:\mathrm{invalid}\: \\ $$$$\mathrm{Q}\:\mathrm{Code}\:−\:\mathrm{42408}\:,\:\mathrm{42471}\:\mathrm{and}\:\mathrm{43496} \\ $$

Commented by LYCON TRIX last updated on 12/Sep/18

I respect your efforts but , we want x  in terms of a , moreover , x ∈ [ −a , +a ]  so be careful

$$\mathrm{I}\:\mathrm{respect}\:\mathrm{your}\:\mathrm{efforts}\:\mathrm{but}\:,\:\mathrm{we}\:\mathrm{want}\:\mathrm{x} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:,\:\mathrm{moreover}\:,\:\mathrm{x}\:\in\:\left[\:−\mathrm{a}\:,\:+\mathrm{a}\:\right] \\ $$$$\mathrm{so}\:\mathrm{be}\:\mathrm{careful} \\ $$

Answered by behi83417@gmail.com last updated on 12/Sep/18

it should be:  a>1.for a<1,there is  no solution.

$${it}\:{should}\:{be}:\:\:{a}>\mathrm{1}.{for}\:{a}<\mathrm{1},{there}\:{is} \\ $$$${no}\:{solution}. \\ $$

Commented by LYCON TRIX last updated on 12/Sep/18

Can you prove it any way ?

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{any}\:\mathrm{way}\:? \\ $$

Commented by behi83417@gmail.com last updated on 12/Sep/18

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