If - - are the roots of a x2-b x-c-0 and - -2-2- -3-3 are in G-P- where -b2-4 ac- then

The correct option is D CONVENTIONAL APPROACH (α^2+β^2)^2=(α+β)(α^3+β^3) (b^2 2ac/a^2)^2=( b/a)( b^2+3abc/a^3) ⇒ 4a^2c^2=acb^2⇒ ac(b^2 4ac)=0 As a0⇒ c =0 ...

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

Byju's Answer

Standard XII

Mathematics

Discriminant

If α, β are t...

Question

If $α, β$ are the roots of $a x^{2} + b x + c = 0$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P., where $△ = b^{2} - 4 a c$ , then

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

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

Open in App

Solution

The correct option is D
$C O N V E N T I O N A L A P P R O A C H$
$(α^{2} + β^{2})^{2} = (α + β) (α^{3} + β^{3})$
${(\frac{b^{2} - 2 a c}{a^{2}})}^{2} = (\frac{- b}{a}) (\frac{- b^{2} + 3 a b c}{a^{3}})$
$\Rightarrow 4 a^{2} c^{2} = a c b^{2} \Rightarrow a c (b^{2} - 4 a c) = 0$
As $a \neq 0 \Rightarrow c △ = 0$

Suggest Corrections

Similar questions

Q. If

α, β

are the roots of

a x^{2} + b x + c = 0 (a \neq 0)

and

α + β, α^{2} + β^{2}, α^{3} + β^{3}

are in

G . P .

, then the value of

c \cdot Δ

is
(where

Δ = b^{2} - 4 a c

)

Q. In the quadratic equation

a x^{2} +

+ c = 0

, if

Δ = b^{2} - 4 a c

and

α + β, α^{2} + {β^{2}, α}^{3} + β^{3}

are in G.P. where

α, β

are the roots of

a x^{2} +

+ c = 0

, then

Q. If

α, β

are the roots of

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

and

α + β, α^{2} + β^{2}, α^{3} + β^{3}

are in G.P., where

△ = b^{2} - 4 a c

, then

Q. In the quadratic equation

a x^{2} + b x + c = 0, Δ = b^{2} - 4 a c a n d α + β, α^{2} + β^{2}, α^{3} + β^{3},

are in G.P. where

α, β

are the root of

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

then

Q. Let

α, β

be roots of the equation

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

and

Δ = b^{2} - 4 a c

. If

α + β, α^{2} + β^{2}, α^{3} + β^{3}

are in

G P

, then

Join BYJU'S Learning Program

If α,β are the roots of ax2+bx+c=0 and α+β,α2+β2,α3+β3 are in G.P., where △=b2−4ac, then

The correct option is D CONVENTIONAL APPROACH (α2+β2)2=(α+β)(α3+β3) (b2−2aca2)2=(−ba)(−b2+3abca3) ⇒4a2c2=acb2⇒ac(b2−4ac)=0 As a≠0⇒c△=0

If $α, β$ are the roots of $a x^{2} + b x + c = 0$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P., where $△ = b^{2} - 4 a c$ , then