Q. The quadratic equation in $x$ such that the arithmetic mean of its roots is 5 and geometric mean of the roots is 4, is given by
A) $x^{2} + 20x + 16 = 0$
B) $x^{2} - 10x + 16 = 0$
C) $x^{2} + 10x + 16 = 0$
D) $x^{2} - 10x - 16 = 0$
Explanation
(b)
Let $\alpha$ and $\beta$ be the roots of the given equation, then
$\alpha + \beta = 10,\ \ \alpha\beta = 16$
$\therefore$ Required equation is
$x^{2} - (\alpha + \beta)x + \alpha\beta = 0$
$\Rightarrow x^{2} - 10x + 16 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.