You visited us 1 times! Enjoying our articles? Unlock Full Access!

Nature of Roots

lf α and ...

Question

lf $α$ and $β$ are the roots of the equation $a x^{2} + b x + c = 0$ and if $p x^{2} + q x + r = 0$ has roots $\frac{1 - α}{α}$ and $\frac{1 - β}{β}$ , then $r =$

a + 2 b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a + b + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a b + b c + c a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a b c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $a + b + c$
$L e t α, β b e t h e r o o t s o f t h e g i v e n e q u a t i o n a x^{2} + b x + c = 0 ∴ α + β = - \frac{b}{a} . . . . (1) a n d α β = \frac{c}{a} . . . . . (2) N o w, t h e e q u a t i o n h a v i n g r o o t s \frac{1 - α}{α}, \frac{1 - β}{β} w i l l b e x^{2} - (\frac{1 - α}{α} + \frac{1 - β}{β}) x + (\frac{1 - α}{α}) (\frac{1 - β}{β}) = 0 \Rightarrow x^{2} - \frac{2 - (α + β)}{α β} x + \frac{1 - (α + β) + α β}{α β} = 0 S u b s t i t u t i n g f o r α + β a n d α β f r o m e q u a t i o n s (1) a n d (2), w e g e t, x^{2} - \frac{2 a + b}{c} x + \frac{a + b + c}{c} = 0 \Rightarrow c x^{2} - (2 a + b) x + (a + b + c) = 0 T h i s e q u a t i o n i s i d e n t i c a l w i t h p x^{2} + q x + r = 0 ∴ c o m p a r i n g t h e l a s t t e r m s, w e g e t, r = a + b + c A n s - O p t i o n B .$

Suggest Corrections

Similar questions

α, β

a x^{2} + b x + c = 0

\frac{1 - α}{α}

\frac{1 - β}{β}

p x^{2} + q x + r = 0

r =

α

β

a x^{2} + b x + c = 0

p x^{2} + q x + r = 0

\frac{1 - α}{α}

\frac{1 - β}{β}

r

α

β

a x^{2} + b x + c = 0.

\frac{1 - α}{α}

\frac{1 - β}{β}

p x^{2} + q x + r = 0

\frac{q}{p} = k + \frac{b}{c}

k

If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0 a n d α + h, β + h$ are the roots of $p x^{2} + q x + r = 0 (h \neq 0)$ , then

α, β

a x^{2} + b x + c = 0

α + h, β + h

p x^{2} + q x + r = 0

h =

Join BYJU'S Learning Program