If - - be the roots of the equation a x 2 -bx-c-0 then the value of -a- 2 b- -c - -a- 2 b- -c is equal to

Here,α and β are roots of the equation a x ^ 2 +bx+c=0 ∴ α +β = b a & αβ = c a ⇒ aα +aβ = b & aαβ =c aα ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Solving a Quadratic Equation by Factorization Method

If α ,β be ...

Question

$I f α, β b e t h e r o o t s o f t h e e q u a t i o n a x^{2} + b x + c = 0 t h e n t h e v a l u e o f \frac{- a α^{2}}{b α + c} + \frac{- a β^{2}}{b β + c} i s e q u a l t o$

Open in App

Solution

Suggest Corrections

Similar questions

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

α

β

a α^{2} + b α + c = 0

a β^{2} + b β + c = 0

α, β

x^{2} - p (x + 1) - c = 0, c \neq 1

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β 2 + 2 β + c} =

α, β

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

\frac{1}{(a α)^{2}} + \frac{1}{(a β)^{2}}

α, β

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

\frac{(a α^{2} + c)}{(a α + b)} + \frac{(a β^{2} + c)}{(a β + b)}

α

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

a α^{2} + b α + c = 0

c \neq 0

α, β

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

\frac{a α^{2} + c}{a α + b} + \frac{a β^{2} + c}{a β + b}

Join BYJU'S Learning Program