How do you find the complex roots of #x^2-14x+49=0#?

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Answer 1 · 2017-01-23T09:58:15

See below.

Assuming complex their roots, the polynomial can be represented as

#(x+a+ib)(x+a-ib)=(x+a)^2+b^2=x^2+2ax+a^2+b^2=0#

Comparing

#{(2a=-14),(a^2+b^2=49):}#

so

#a=-7# and #b=0#

so the polynomial is #(x+7)^2=0# with real roots.