Q. $\sqrt{4\ },\sqrt[4]{4},\ \sqrt[8]{4},\ \sqrt[16]{4},\ \ldots to\ \infty$ are root of equation
A) $x^{2} - 4 = 0$
B) $x^{2} - 4x + 6 = 0$
C) $x^{2} - 5x + 4 = 0$
D) $x^{2} - 3x + 2 = 0$
Explanation
(c)
$4^{1/2,},\ 4^{1/4},\ 4^{1/8},\ 4^{1/16},\ \ldots$ are given roots, then
Sum of roots$= 4^{\frac{1}{2}} + 4^{\frac{1}{4}} + 4^{\frac{1}{8}} + \ldots = 5$
Product of roots$= 4^{\frac{1}{2}}.4^{\frac{1}{4}}.4^{1/8}\ldots$
$= 4^{1/2 + 1/4 + 1/8 + \ldots}$
$= 4^{\frac{1/2}{1 - 1/2}} = 4$
$\therefore$ Required equation is $x^{2} - 5x + 4 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.