$\begin{array}{l} I f α, β a r e t h e r o o t s o f t h e e q u a t i o n x^{2} + 2 a x + b = 0, t h e n t h e q u a d r a t i c e q u a t i o n w i t h r a t i o n a l c o e f f i c i e n t s, o n e o f w h o s e \\ r o o t s i s α + β + \sqrt{α^{2} + β^{2}} i s \end{array}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x^{2} - 4 a x + 2 b = 0$

$x^{2} + 4 a x - 2 b = 0$

$x^{2} - 4 a x - 2 b = 0$

$x^{2} + 4 a x + 2 b = 0$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I f α, β a r e r o o t s x^{2} + 2 a x + b = 0 \\ α + β = - 2 a, α β = b \\ ∴ α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = 4 a^{2} - 2 b \\ L e t γ = (α + β) + \sqrt{α^{2} + β^{2}} t h e n δ = (α + β) - \sqrt{α^{2} + β^{2}} \\ γ + δ = 2 (α + β) = - 4 a \\ γ δ = {(α + β)}^{2} - (α^{2} + β^{2}) = 2 α β = 2 b \\ Re q u i r e d E q u a t i o n i s x^{2} - (γ + δ) x + γ δ = 0 \\ \Rightarrow x^{2} + 4 a x + 2 b = 0 \end{array}$