Conditiile de existenta sunt 78+x >= 0 <=> x >= -78 si respectiv 19-x >= 0 <=>
x <= 19 <=> x∈[-78,19] .
⁴√(78+x) +⁴√(19-x) =5 <=> notam ⁴√(78+x) =a si ⁴√(19-x) =b => a+b=5 si a⁴+b⁴=97 => a⁴+(5-a)⁴=97 => a=3 => b=2 .
Asadar 78+x=81 si 19-x=16 => x=3 .