JEE PYQ: Quadratic Equation Question 11
Question 11 - 2021 (25 Feb Shift 2)
If $\alpha, \beta \in \mathbf{R}$ are such that $1 - 2i$ (here $i^2 = -1$) is a root of $x^2 + \alpha x + \beta = 0$, then $(\alpha - \beta)$ is equal to:
(1) 7
(2) -3
(3) 3
(4) -7
Show Answer
Answer: (4)
Solution
$(1 - 2i)^2 + \alpha(1 - 2i) + \beta = 0$
$1 - 4 - 4i + \alpha - 2i\alpha + \beta = 0$
$(\alpha + \beta - 3) - i(4 + 2\alpha) = 0$
$\alpha + \beta - 3 = 0$ & $4 + 2\alpha = 0$
$\alpha = -2$, $\beta = 5$
$\alpha - \beta = -7$
BETA
