If each pair of the following three equations $x^{2} + a x + b = 0$ , $x^{2} + c x + d = 0$ , $x^{2} + e x + f = 0$ has exactly one root in common, then

{(a + c + e)}^{2} = 3 (a c + c e + e a - b - d - f)

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{(a + c + e)}^{2} = 8 (a c + c e + e a - b - d - f)

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{(a + c + e)}^{3} = 4 (a c + c e + e a - b - d - f)

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{(a + c + e)}^{2} = 4 (a c + c e + e a - b - d - f)

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Solution

The correct option is D ${(a + c + e)}^{2} = 4 (a c + c e + e a - b - d - f)$
Given equation are
$x^{2} + a x + b = 0$ ........(1)
$x^{2} + c x + d = 0$ ...........(2)
$x^{2} + e x + f = 0$ ..........(3)
Let $α, β$ be the roots of (1), $β, γ$ be the roots of (2) and $γ, α$ be the roots of (3)
$∴ α + β = - a, α β = b$ ...........(4)
$β + γ = - c, β γ = d$ ............ (5)
$γ + α = - e, γ α = f$ .........(6)
$∴$ L.H.S. $= {(a + c + e)}^{2}$
$= {(- α - β - β - γ - γ - α)}^{2}$
{from (4),(5),& (6)}
$= {(α + β + γ)}^{2}$ ........(7)
R.H.S = $4 (a c + c e + e a - b - d - f)$
$= 4 {(α + β) (β + γ) + (β + γ) (γ + α) + (γ + α) (α + β) - α β - β γ - γ α}$
{from (4), (5)& (6)}
$= 4 (α^{2} + β^{2} + γ^{2} + 2 α β + 2 β γ + 2 γ α)$
$= 4 {(α + β + γ)}^{2}$ .....(8)
From (7) & (8),
$= {(a + c + e)}^{2} = 4 (a c + c e + e a - b - d - f)$ .

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