• # question_answer If $a<b<c<d$, then the roots of the equation $(x-a)(x-c)+2(x-b)(x-d)=0$ are [IIT 1984] A) Real and distinct B) Real and equal C) Imaginary D) None of these

Given equation can be rewritten as $3{{x}^{2}}-(a+c+2b+2d)x+(ac+2bd)=0$ Its discriminant D $={{(a+c+2b+2d)}^{2}}-4.3(ac+2bd)$ $={{\left\{ (a+2d)+(c+2b) \right\}}^{2}}-12(ac+2bd)$      $={{\left\{ (a+2d)-(c+2b) \right\}}^{2}}+4(a+2d)(c+2b)-12(ac+2bd)$     $={{\left\{ (a+2d)-(c+2b) \right\}}^{2}}-8ac+8ab+8dc-8bd$ $={{\left\{ (a+2d)-(c+2b) \right\}}^{2}}+8(c-b)(d-a)$ which is +ve, since $a<b<c<d$. Hence roots are real and distinct.