Question

(x^2-49)^2 + (x^2+4x-21)^2 = 0

Answer 1

A²≥0, a∈(-∞,∞)
(x²-49)²+(x²+4x-21)²=0 ⇔ (x²-49)=0 ∧ (x²+4x-21)=0
a)x²-49=0, (x+7)(x-7)=0, x=-7,x=7
b)x²+4x-21=0,D=16+84=100, √D=√100=10
  x=(-4+10)/2=6/2=3
  x=(-4-10)/2=-14/2=-7
Otvet: x=-7, x=3, x=7