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If a < b < c < d, then the roots of the equation (x-a)(x -c) +2(x - b)(x - d) =0 are

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a

real and distinct

b

real and equal

c

imaginary

d

None of these

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detailed solution

Correct option is A

Given equation can be rewritten as
$3 x^{2} - (a + c + 2 b + 2 d) x + ac + 2 bd = 0$
$∴$ Discriminant, D
$\begin{array}{l} = (a + c + 2 b + 2 d)^{2} - 4 \cdot 3 (a c + 2 b d) \\ = {(a + 2 d) + (c + 2 b)}^{2} - 12 (a c + 2 b d) \\ = {(a + 2 d) - (c + 2 b)}^{2} + 4 (a + 2 d) (c + 2 b) - 12 (a c + 2 b d) \\ = {(a + 2 d) - (c + 2 b)}^{2} - 8 a c + 8 a b + 8 d c - 8 b d \\ = {(a + 2 d) - (c + 2 b)}^{2} + 8 (c - b) (d - a) \end{array}$
Which is +ve, since a < b < c < d.
Hence, roots are real and distinct.

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