(x^2-4)^2+(x^2-6x-16)^2=0

Question

(x^2-4)^2+(x^2-6x-16)^2=0

Answer 1

(x^2-4)^2+(x^2-6x-16)^2=0
2x^4-12x^3-4x^2+192x+272=0
2(x^4-6x^3-2x^2+96x+136)=0
2(x+2)^2(x^2-10x+34)=0
(x+2)^2=0 => x+2=0 => x=-2
или:
x^2-10x+34=0
D=-10^2-4*1*34=-36
Нет вещественных корней
Ответ: х=-2