# Thread: need to solve this problem

1. ## need to solve this problem

a<b<c<d
f(x)=(x-a)(x-c)-3(x-b)(x-d)
show that the roots of this equation are distinct.

2. 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.

3. Alternatively,

since f(c)<0 and f(d)>0, the roots of the quadratic are real and distinct since the graph crosses the x-axis.

The roots are non-distinct in the case of a double-root whereby the graph does not cross the x-axis.

4. what if the question is $\displaystyle a<b<c<d$
$\displaystyle f(x)=(x-a)(x-c)+(x-b)(x-d)$
does it have two distinct roots??