# If $a,b,c \in R$ such that $a < b < c < d$ , show that $(x-a)(x-c) + 2(x-b)(x-d) = 0$ has real and distinct roots . [duplicate]

If $$a,b,c \in R$$ such that $$a < b < c < d$$ , show that $$(x-a)(x-c) + 2(x-b)(x-d) = 0$$ has real and distinct roots .

I'm not getting good manipulation idea to show discriminant positive , also wherever I searched i found only same not much intuitive solution . I m seeking for good intuitive solutions here thanks in advance.

Someone shared a neat idea of sign changes here but still I'm looking for some good manipulative ideas too .

• The tag (theory-of-equations) was previously discussed on meta and the consensus was that the tag should be removed. I'd say that if you still think that such tag should exist, you should probably start a discussion on meta about that tag - to see whether other users of the site consider such tag to be useful. Commented Jun 17 at 10:58
• @MartinSleziak i didn't know about it sorry but I m fine with removal of this tag Commented Jun 17 at 11:39

If $$f(x)=(x-a)(x-c) + 2(x-b)(x-d)$$ then $$f(a)>0$$, $$f(b)<0$$ and $$f(d)>0$$, so there must be a root between $$a$$ and $$b$$ and a root between $$b$$ and $$d$$.