JEE PYQ: Quadratic Equation Question 27
Question 27 - 2020 (08 Jan Shift 1)
If the equation, $x^2 + bx + 45 = 0$ $(b \in R)$ has conjugate complex roots and they satisfy $|z + 1| = 2\sqrt{10}$, then:
(1) $b^2 - b = 30$
(2) $b^2 + b = 72$
(3) $b^2 - b = 42$
(4) $b^2 + b = 12$
Show Answer
Answer: (1)
Solution
Let $z = \alpha \pm i\beta$ be the complex roots of the equation.
So, sum of roots $= 2\alpha = -b$ and product of roots $= \alpha^2 + \beta^2 = 45$
$(\alpha + 1)^2 + \beta^2 = 40$ (wait: $|z+1| = 2\sqrt{10}$ so $|z+1|^2 = 40$)
Given, $|z + 1| = 2\sqrt{10}$
$\Rightarrow (\alpha + 1)^2 + \beta^2 = 40$ (but $\alpha^2 + \beta^2 = 45$)
$\Rightarrow (\alpha + 1)^2 - \alpha^2 = -5 \Rightarrow 2\alpha + 1 = -5 \Rightarrow 2\alpha = -6$
Hence, $b = 6$ and $b^2 - b = 30$
BETA
