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

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