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

$a<b<c<d$

Roots of the equation

$(x-a)(x-c)+2(x-b)(x-d)=0$

$3x^2-x(a+c+2b+2d)+ac+2bd=0$

$\Delta =b^2-4ac$

$={(a+c+2b+2d)}^2-\mathrm{4.3.}(ac+2bd)$

$=a^2+c^2+2ac+4b^2+4d^2+8bd+2(a+c)(2b+2d)-12ac-24bd$

$=a^2+c^2+4b^2+4d^2-10ac+8bd+4ab+4ad+4cb+4cd-24bd$

$=a^2+c^2+4b^2+4d^2-10ac+4ab+4ad+4bc+4cd-16bd$

$=\frac{1}{4}[4a^2+4c^2+16b^2+16d^2-40ac+16ab+16ad+16bc+16cd-64bd]$

$\Delta >0$ Roots are real and distinct.