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

The correct option is B 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 a 0 ⇒ c =0 T ...

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Byju's Answer

Standard XII

Mathematics

Sum of Infinite Terms of a GP

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

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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 $a \neq 0 \Rightarrow c △ = 0$
Trick : Assume $α = 0, β = 2$
Then, $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ i.e. 2,4,6 are in GP.
And the equation is reduced to $x (x - 2) = x^{2} - 2 x$
Therefore, a=1, b=-2, c=0
Only option (d) satisfies.

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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

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