Question

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

Answer 1

$(x^2-49)^2+(x^2+4x-21)^2=0\(x-7)^2(x+7)^2+(x^2+7x-3x-21)^2=0\(x-7)^2(x+7)^2+(x(x-3)+7(x-3))^2=0\(x-7)^2(x+7)^2+(x+7)^2(x-3)^2=0\(x+7)^2((x-7)^2+(x-3)^2)=0\\x+7=0\x=-7;\\(x-7)^2+(x-3)^2=0\x^2-14x+49+x^2-6x+9=0\2x^2-20x+58=0|:2\x^2-10x+29=0\D: ; 100-116=-16 textless 0.$