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Q.

If $(α, β), (β, γ) a n d (γ, α)$ are respectively the roots of $x^{2} - 2 p x + 2 = 0, x^{2} - 2 q x + 3 = 0 a n d x^{2} - 2 r x + 6 = 0$ , where $α, β, γ$ are all positive, then the value of $p + q + r$ is

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answer is 6.

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Detailed Solution

$\begin{array}{l} α + β = 2 p \\ β + γ = 2 q \\ \underline{γ + α = 2 γ} \\ 2 (α + β + γ) = 2 (p + q + r) \\ α + β + γ = p + q + r \\ α β = 2, β γ = 3, γ α = 6 \\ α^{2} = \frac{α β \times γ α}{β γ} = \frac{2 \times 6}{3} = 4 \\ β^{2} = \frac{α β \times β γ}{α γ} = \frac{2 \times 3}{6} = 1 \Rightarrow β = 1 \\ γ^{2} = \frac{γ α + β γ}{α β} = \frac{6 \times 3}{2} = 9 \\ γ = 3 \\ ∴ p + q + r = α + β + γ = 2 + 1 + 3 = 6 \end{array}$

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If (α,β),(β,γ) and (γ,α) are respectively the roots of x2−2px+2=0,x2−2qx+3=0 and x2−2rx+6=0, where α,β,γ are all positive, then the value of p+q+r is