Question

If $\alpha ,\beta \in R$ are such that $1–2i(i^2=–1)$ is a root of $z^2+\alpha z+\beta =0$, then $(\alpha –\beta )$ is equal to:

• A. $7$
• B. $-3$
• C. $3$
• D. $-7$

Answer 1

The correct option is D

$-7$

Find the value of $\mathbf{(}\mathit{\alpha }\mathbf{}\mathbf{–}\mathbf{}\mathit{\beta }\mathbf{)}$:

Given, $1-2i$ is a root of $z^2+\alpha z+\beta =0$.

$\Rightarrow$ ${(1-2i)}^2+\alpha (1-2i)+\beta =0$

$\Rightarrow$ $1-4-4i+\alpha -2i\alpha +\beta =0$

$\Rightarrow$ $(\alpha +\beta -3)-i(4+2\alpha )=0$

$\Rightarrow$ $\alpha +\beta -3=0,4+2\alpha =0$

$\Rightarrow$ $\alpha +\beta -3=0,\alpha =-\frac{4}{2}$

$\Rightarrow$ $\alpha +\beta =3,\alpha =-2$

$\Rightarrow$ $-2+\beta =3,\alpha =-2$

$\Rightarrow$ $\beta =3+2,\alpha =-2$

$\Rightarrow$ $\alpha =-2,\beta =5$

$\mathbf{\therefore }\mathit{\alpha }\mathbf{}\mathbf{-}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{7}$

Hence, Option ‘D’ is Correct.