题目内容
解关于x的方程
①x2+2x+3=0;②x2+6x+13=0.
①x2+2x+3=0;②x2+6x+13=0.
①x=-1+i或x=-1-i②x=-3+2i或x=-3-2i
①设x=a+bi(a,b∈R),
则x2+2x+3=a2-b2+2abi+2a+2bi+3
=(a2-b2+2a+3)+(2ab+2b)i=0.
∵a,b∈R,∴a2-b2+2a+3=0且2ab+2b=0.
∴或 ∴x=-1+i或x=-1-i
②设x=a+bi(a,b∈R),
则x2+6x+13=a2-b2+2abi+6a+6bi+13
=a2-b2+6a+13+(2ab+6b)i=0.
∵a,b∈R,∴a2-b2+6a+13=0且2ab+6b=0.
∴或 ∴x=-3+2i或x=-3-2i
则x2+2x+3=a2-b2+2abi+2a+2bi+3
=(a2-b2+2a+3)+(2ab+2b)i=0.
∵a,b∈R,∴a2-b2+2a+3=0且2ab+2b=0.
∴或 ∴x=-1+i或x=-1-i
②设x=a+bi(a,b∈R),
则x2+6x+13=a2-b2+2abi+6a+6bi+13
=a2-b2+6a+13+(2ab+6b)i=0.
∵a,b∈R,∴a2-b2+6a+13=0且2ab+6b=0.
∴或 ∴x=-3+2i或x=-3-2i
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