If 1-i is a root of the equation x^2+ax+b=0 w | vedclass

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

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$2$

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(D) Since the coefficients $a$ and $b$ are real,the complex roots of the quadratic equation $x^2+ax+b=0$ must occur in conjugate pairs. Given that $1-i$ is a root,its conjugate $1+i$ must also be a root of the equation. For a quadratic equation $x^2+ax+b=0$,the product of the roots is given by the constant term $b$. Therefore,$b = (1-i)(1+i)$. Using the identity $(x-y)(x+y) = x^2-y^2$: $b = 1^2 - i^2$. Since $i^2 = -1$,we have: $b = 1 - (-1) = 1 + 1 = 2$.

If $1-i$ is a root of the equation $x^2+ax+b=0$ where $a$ and $b$ are real numbers,then $b$ is equal to

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