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.

Question

How do I prove the above problem?

I tried finding discriminant, It wasn’t very useful as it didn’t yield any useful conclusion.

Answer 1

Note that

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

$$\iff g(x)=(x-a)(x-c)=-2(x-b)(x-d)=h(x)$$

and

• $g(x)$ is a parabola concave up which intercepts $x$ axis at $x=a$ and $x=c$
• $h(x)$ is a parabola concave down which intercepts $x$ axis at $x=b$ and $x=d$

thus by IVT $2$ distinct real roots exist for $f(x)=0$.