Position of roots of $(x-a)(x-b)+2(x-c)(x-d)=0.$ Problem: If $a<b<c<d$, then consider the equation $(x-a)(x-b)+2(x-c)(x-d)=0.$ Does the equation have both roots in $[a,b]$, both roots in $[c,d]$ or have one root in $(a,b)$ and the other root in $(c,d)$?
I assume the equation must have real roots for the problem to make sense.
My approach: Differentiating the quadratic function, the minimum value of the quadratic function is obtained at $$x=\frac{a+b+2c+2d}{6}\in (a,d).$$ Since, the parabola is concave up and $f(a),f(d)>0$, both roots must lie in $(a,d)$. Since $f(b)>0,f(c)>0$, the roots must lie in $(a,b)$ or in $(c,d)$.
How do I proceed from here?
 A: The function:
$$ f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} $$
is at most $2$-to-$1$ and has two singularities at $x=c$ and $x=d$. In order to know where the solutions of $f(x)=-2$ lie, we just need to locate the stationary points of $f(x)$ and compute the values of $f(x)$ in such points. The stationary points are given by the solutions of:
$$ (a+b-c-d)x^2 +2(cd-ab) x + abcd\left(\frac{1}{c}+\frac{1}{d}-\frac{1}{a}-\frac{1}{b}\right)=0$$
one of them (the relative minimum, $x_m$) belonging to $(a,b)$ and the other one (the relative maximum, $x_M$) belonging to $(c,d)$ by Rolle's theorem. About the solutions of $f(x)=-2$: if $f(x_m)\leq -2$, they both belong to $(a,b)$ - it is the case for $(a,b,c,d)=(1,100,101,102)$, for instance. If $f(x_M)\geq -2$, they both belong to $(c,d)$: it is the case for $(a,b,c,d)=(1,2,3,10)$, for instance. If neither of the two conditions are fulfilled, there are no solutions: it is the case for $(a,b,c,d)=(1,2,3,4)$, for instance.
A: Since  $a<b<c<d $. 
$F(x)>0$ when $x<a$,
$F(x)<0$ when $a<x<b$,
$F(x)>0$ when $b<x<c$,
$F(x)<0$ when $c<x<d$,
$F(x)>0$ when $x>d$.
And $F(a)=F(b)=0$.
 And $F(c)\to-\infty$ from right of $c$.
And $F(d)\to+\infty$ from right of $d$.
$F(x)=-2$ only in interval of $(a,b)$ or $(c,d)$.
Therefore roots lies in two intervals.
