The equation of the smallest degree with real | vedclass

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(B) Since the coefficients are real,if $1 + i$ is a root,then its conjugate $1 - i$ must also be a root.Let the roots be $\alpha = 1 + i$ and $\beta =

Vedclass · Accepted Answer

$x^2 - 2x + 2 = 0$

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(B) Since the coefficients are real,if $1 + i$ is a root,then its conjugate $1 - i$ must also be a root.Let the roots be $\alpha = 1 + i$ and $\beta = 1 - i$.The sum of the roots is $\alpha + \beta = (1 + i) + (1 - i) = 2$.The product of the roots is $\alpha \beta = (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - 2x + 2 = 0$.

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

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