Question

(x^2-4)^2+(x^2+4x-12)^2=0 как решить?

Answer 1

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