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Question Number 73131 by behi83417@gmail.com last updated on 06/Nov/19

solve for x,in terms of: a∈R . x+(√x)+(√(x^2 −a))+(√(x−a^2 ))=a^2

solveforx,intermsof:a∈R.x+x+x2−a+x−a2=a2

Commented by mr W last updated on 06/Nov/19

x≥0 x−a^2 ≥0 x^2 −a≥0 (x−a^2 )+(√x)+(√(x^2 −a))+(√(x−a^2 ))≥0 ⇒only solution is, when x−a^2 =0 x=0 x^2 −a=0 ⇒a=0 and x=0 if a≠0, there is no solution.

x⩾0x−a2⩾0x2−a⩾0(x−a2)+x+x2−a+x−a2⩾0⇒onlysolutionis,whenx−a2=0x=0x2−a=0⇒a=0andx=0ifa≠0,thereisnosolution.

Commented by behi83417@gmail.com last updated on 06/Nov/19

thanks in advance dear master.

thanksinadvancedearmaster.

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