If 1 - i is a root of the equation {x^2} - ax | vedclass

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Question

(D) Since the coefficients of the quadratic equation ${x^2} - ax + b = 0$ are real,the complex roots must occur in conjugate pairs. Given that $1 - i$

Vedclass · Accepted Answer

$ 2 $

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(D) Since the coefficients of the quadratic equation ${x^2} - ax + b = 0$ are real,the complex roots must occur in conjugate pairs. Given that $1 - i$ is a root,its conjugate $1 + i$ must also be a root. The sum of the roots is $(1 - i) + (1 + i) = 2$. From the equation,the sum of the roots is $a$,so $a = 2$. The product of the roots is $(1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2$. From the equation,the product of the roots is $b$,so $b = 2$.

If $1 - i$ is a root of the equation ${x^2} - ax + b = 0$,then $b = $

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