The equation of the smallest degree with real coefficients having 1+i as one of the roots is

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x2+x+1=0

x2−2x+2=0

x2+2x+2=0

x2+2x−2=0

answer is B.

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Detailed Solution

Let x=1+i ⇔ x−1 =i Squaring on both sides,we get ⇔ (x−1)2=i2 ⇒ x2−2x+1 =−1 ⇔ x2−2x+2=0.

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