a<b<c<d
f(x)=(x-a)(x-c)-3(x-b)(x-d)
show that the roots of this equation are distinct.
The simplest way to do this is to look at the values of f(a), f(b), f(c), f(d)
f(a)= -3(a- b)(a- d). Since a< b and a< d, both a- b and a- d are less than 0 so f(a)< 0.
f(b)= (b- a)(b- c). a< b but b< c so b- a is positive, b- c is negative so f(b)< 0
f(c)= -2(c- b)(c- d). b< c but c< d so c- b is positive and c-d is negative so f(c)< 0
f(d)= (d- a)(d- b). a< d and b< d so both d- a and d- b are positive. f(d)> 0 so there is a root between c and d.
Multiplying out, $\displaystyle f(d)= x^2- (a+ c)x+ ac- 3x^2+ 3(b+d)x- 3bd= -2x^2- (a+ c- b- d)x+ ac- 3bd$
Since the leading coefficient is negative, for sufficiently large x, f(x) will be negative. There must be a second root larger than d.