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Question Number 62228 by behi83417@gmail.com last updated on 17/Jun/19

{ (((√(a+x))+(√(a−y))=2a)),(((√(a−x))+(√(a+y))=2a)) :} a∈R.

{a+x+a−y=2aa−x+a+y=2aa∈R.

Commented by mr W last updated on 18/Jun/19

a>0 −a≤x,y≤a x=y (√(a+x))+(√(a−x))=2a (√(a+x))=2a−(√(a−x)) a+x=4a^2 −4a(√(a−x))+a−x x=2a^2 −2a(√(a−x)) 2a^2 −x=2a(√(a−x)) 4a^4 −4a^2 x+x^2 =4a^3 −4a^2 x x^2 =4(1−a)a^3 ⇒x=y=±2a(√(a(1−a))) ⇒a≤1 2a(√(a(1−a)))≤a ⇒a≥(1/2) i.e. solution exists only when (1/2)≤a≤1.

a>0−a⩽x,y⩽ax=ya+x+a−x=2aa+x=2a−a−xa+x=4a2−4aa−x+a−xx=2a2−2aa−x2a2−x=2aa−x4a4−4a2x+x2=4a3−4a2xx2=4(1−a)a3⇒x=y=±2aa(1−a)⇒a⩽12aa(1−a)⩽a⇒a⩾12i.e.solutionexistsonlywhen12⩽a⩽1.

Commented by behi83417@gmail.com last updated on 18/Jun/19

thank you very much dear master.

thankyouverymuchdearmaster.

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