QJEEM20Ma8fb8: JEE Main 2020 Mathematics - Class 11 JEEM20Ma8fb8 Solution: If the equation, x 2 + bx + 45 = 0 (b ∈ R) has conjugate…

x² + bx = 45 = 0 (b $$\in$$ R)
has roots $$\alpha$$ + i$$\beta$$, $$\alpha$$ – i$$\beta$$
sum of roots = – b = 2$$\alpha$$
product of roots = 45 = $$\alpha$$² + $$\beta$$²

Let z = x + iy

$$\therefore$$ |x + iy +1| = 2$$\sqrt{10}$$

$${\left|x+iy+1\right|}^2={\left(2\sqrt{10}\right)}^2$$

$$\Rightarrow$$ (x + 1)² + y² = 40

$$\Rightarrow$$ ($$\alpha$$ + 1)² + $$\beta$$² = 40

[putting real part $$\alpha$$ in place of x and imaginary part $$\beta$$ in place of y]

$$\Rightarrow$$ $$\alpha$$² + 2$$\alpha$$ + 1 + $$\beta$$² = 40

$$\Rightarrow$$ 45 + 2$$\alpha$$ + 1 = 40

$$\Rightarrow$$ $$\alpha$$ = -3

$$\therefore$$ -b = 2$$\alpha$$ = 2$$\times$$(-3) = -6

$$\Rightarrow$$ b = 6

By checking options we found b² – b = 30.