The harmonic mean of two numbers is $- \frac{8}{5}$ and their geometric mean is 2. The quadratic equation whose roots are twice those numbers, is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$x^{2} - 10 x + 16 = 0$

$x^{2} + 5 x + 4 = 0$

$x^{2} - 5 x + 4 = 0$

$x^{2} + 10 x + 16 = 0$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} L e t a a n d b a r e t w o n u m b e r s \\ G i v e n \frac{2 a b}{a + b} = \frac{- 8}{5} a n d \sqrt{a b} = 2 \Rightarrow a b = 4 \\ ∴ \frac{2 (4)}{a + b} = - \frac{8}{5} \\ \Rightarrow a + b = - 5 \\ L e t α = 2 a, β = 2 b a r e r o o t s \\ \Rightarrow Re q u i r e d E q u a t i o n i s \\ \Rightarrow x^{2} - (2 a + 2 b) x + 2 a (2 b) = 0 \\ \Rightarrow x^{2} - 2 (a + b) x + 4 a b = 0 \\ \Rightarrow x^{2} - 2 (- 5) x + 4 (4) = 0 \\ \Rightarrow x^{2} + 10 x + 16 = 0 \end{array}$