Question:
If $\alpha, \beta \in \mathrm{R}$ are such that $1-2 \mathrm{i}$ (here $\mathrm{i}^{2}=-1$ ) is a root of $\mathrm{z}^{2}+\alpha \mathrm{z}+\beta=0$, then $(\alpha-\beta)$ is equal
to:
Correct Option: , 4
Solution:
$(1-2 i)^{2}+\alpha(1-2 i)+\beta=0$
$1-4-4 i+\alpha-2 i \alpha+\beta=0$
$(\alpha+\beta-3)-i(4+2 \alpha)=0$
$\alpha+\beta-3=0 \quad \& 4+2 \alpha=0$
$\alpha=-2 \quad \beta=5$
$\alpha-\beta=-7$
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