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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
$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\:. \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\sqrt{\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} \\ $$
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.
$${x}\geqslant\mathrm{0} \\ $$$${x}−{a}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{a}\geqslant\mathrm{0} \\ $$$$\left({x}−{a}^{\mathrm{2}} \right)+\sqrt{{x}}+\sqrt{{x}^{\mathrm{2}} −{a}}+\sqrt{{x}−{a}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$$\Rightarrow{only}\:{solution}\:{is},\:{when} \\ $$$${x}−{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{a}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{0}\:{and}\:{x}=\mathrm{0} \\ $$$${if}\:{a}\neq\mathrm{0},\:{there}\:{is}\:{no}\:{solution}. \\ $$
Commented by behi83417@gmail.com last updated on 06/Nov/19
thanks  in advance dear master.
$$\mathrm{thanks}\:\:\mathrm{in}\:\mathrm{advance}\:\mathrm{dear}\:\mathrm{master}. \\ $$

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