Q.

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

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a

x2+x+1=0

b

x2−2x+2=0

c

x2+2x+2=0

d

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