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a-a-x-a-a-x-2x-Solve-for-x-in-terms-of-a-




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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