In the quadratic equation a x 2 + b x + c = 0 , = b 2 - 4 a c and 2 + 2 , 3 + 3 are in G.P. where are the roots of a x 2 + b x + c = 0 , then [2005]

Question

In the quadratic equation $a x^{2} + b x + c = 0, Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P. where $α, β$ are the roots of $a x^{2} + b x + c = 0,$ then [2005]

Smart Achievers · Accepted Answer

(3)

$In the quadratic equation a x^{2} + b x + c = 0$

$Δ = b^{2} - 4 a c and α + β = - \frac{b}{a}, α β = \frac{c}{a}$

$α^{2} + β^{2} = {(α + β)}^{2} - 2 α β$

$= \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{b^{2} - 2 a c}{a^{2}}$

$and α^{3} + β^{3} = - \frac{b^{3}}{a^{3}} - \frac{3 c}{a} (- \frac{b}{a}) = - (\frac{b^{3} - 3 a b c}{a^{3}})$

$Since α + β, α^{2} + β^{2} and α^{3} + β^{3} are in G.P.$

$∴$ $- \frac{b}{a}, - \frac{b^{2} - 2 a c}{a^{2}}, - \frac{b^{3} - 3 a b c}{a^{3}} are in G.P.$

$\Rightarrow {(\frac{b^{2} - 2 a c}{a^{2}})}^{2} = \frac{b}{a} (\frac{b^{3} - 3 a b c}{a^{3}})$

$\Rightarrow b^{4} + 4 a^{2} c^{2} - 4 a b^{2} c = b^{4} - 3 a b^{2} c$

$\Rightarrow 4 a^{2} c^{2} - a b^{2} c = 0 \Rightarrow a c Δ = 0$

$\Rightarrow c Δ = 0$ $(∵ in quadratic equation a \neq 0)$

Solution

Q.
In the quadratic equation $a x^{2} + b x + c = 0, Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P. where $α, β$ are the roots of $a x^{2} + b x + c = 0,$ then [2005]

1 $Δ \neq 0$

2 $b Δ = 0$

3 $c Δ = 0$

4 $Δ = 0$

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Solution

Q. In the quadratic equation ax2+bx+c=0, Δ=b2-4ac and α+β, α2+β2, α3+β3 are in G.P. where α,β are the roots of ax2+bx+c=0, then [2005]

1 Δ≠0

2 bΔ=0

3 cΔ=0