Question

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

Answer 1

Let $f(x)=(x-a)(x-c)+2(x-b)(x-d)$
Here, $f(a)=$ + ve
$f(b)=$- ve
$f(c)=$ - ve
$f(d)=$ + ve
$\therefore$ There exists two real and distinct roots one in the interval $(a,b)$ and other in $(c,d)$.
Hence, statement is true.