Q. The equation $x^{2} - 2\sqrt{2}kx + 2e^{2\log k} - 1 = 0$ has the product of roots equal to 31, then for what value of $k$ it has real roots?
A) 1
B) 2
C) 3
D) 4
Explanation
(d)
Given equation is
$x^{2} - 2\sqrt{2}kx + 2e^{2\log k} - 1 = 0$
Also, product of its root $2e^{2\log k} - 1 = 31$
$\Rightarrow \ \ \ 2\ e^{2\log k} = 32\ \ \ \Rightarrow \ \ \ k^{2} = 16$
$\Rightarrow \ \ k = \pm 4$ [Since, log is not defined for $k < 0$]
$\therefore\ \ \ k = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.