Question

If $1-i$ is a root of the equation $x^2+ax+b=0$, where $a,b\in R$, then the values of $a$ and $b$ are $\underset{–}{}$

Answer 1

As the coefficients of the quadratic equation are real, so The complex roots exist in conjugate pair, as one root is $(1-i)$ another root is $(1+i)$

Sum of roots $=(1+i)+(1-i)=2$

Product of roots

$=(1+i)(1-i)$

$=1-i^2=1-(-1)=2$

The required equation is

$x^2-$(Sum of roots)$x+$ Product of roots $=0$

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

Comparing this with given equation:

$x^2+ax+b=0$, we get

$a=-2,b=2$