If the roots of $(x - α) (x - 4 + β) + (x - 2 + α) (x + 2 - β) = 0$ are $p$ and $q$ , then the value of sum of the roots of $2 (x - p) (x - q) - (x - α) (x - 4 + β) = 0$ and $2 (x - p) (x - q) - (x - 2 + α) (x + 2 - β) = 0$ is

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Solution

$(x - α) (x - 4 + β) + (x - 2 + α) (x + 2 - β) = 2 (x - p) (x - q)$

Now,
$2 (x - p) (x - q) - (x - α) (x - 4 + β) = (x - 2 + α) (x + 2 - β)$
So, roots are $(2 - α), (β - 2)$
$2 (x - p) (x - q) - (x - 2 + α) (x + 2 - β) = (x - α) (x - 4 + β)$
So, the roots are $α, 4 - β$

Sum of roots of both the equation
$= 2 - α + β - 2 + α + 4 - β = 4$

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