Question

(x2-36)2 + (x2+4x - 12)2=0​

Answer 1

$(x^2-36)^2+(x^2+4x-12)^2=0\\\\x^2+4x-12=0\; \; \to \; \; x_1=-6\; ,\; x_2=2\; \; (teorema\; Vieta)\; \; \Rightarrow \\\\(x-6)^2(x+6)^2+(x+6)^2(x-2)^2=0\\\\(x+6)^2\cdot \Big((x-6)^2+(x-2)^2)\Big)=0\\\\a)\; \; (x+6)^2=0\; \; \to \; \; x=-6\\\\b)\; \; (x-6)^2+(x-2)^2=0\\\\x^2-12x+36+x^2-4x+4=0\\\\2x^2-16x+40=0\\\\x^2-8x+20=0\; \; ,\; \; D=-16<0\; \; \Rightarrow \; \; \; x\in \varnothing \\\\Otvet:\; \; \x=-6\; .$