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

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Answer 1 · 2017-01-23T09:58:15

See below.

Assuming complex their roots, the polynomial can be represented as

$(x + a + i b) (x + a - i b) = {(x + a)}^{2} + b^{2} = x^{2} + 2 a x + a^{2} + b^{2} = 0$ $(x+a+ib)(x+a-ib)=(x+a)^2+b^2=x^2+2ax+a^2+b^2=0$

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.

How do you find the complex roots of x^2-14x+49=0x2−14x+49=0x^2-14x+49=0?