Let a,b,c be real numbers. Prove that three roots of the equation
b+cx−a+c+ax−b+a+bx−c=3
are real.
Solution: The solution which the book lists is actually much more elegant than what I will present here. We let s=a+b+c to be the sum. Inserting this into the equation, we see that
s−ax−a+s−bx−b+s−cx−c=3
Now, multiplying both sides by (x−a)(x−b)(x−c) and rearranging results in
(s−a)(x−b)(x−c)+(s−b)(x−a)(x−c)+(s−c)(x−a)(x−b)–3(x−a)(x−b)(x−c)=0.
By inspection, it’s pretty clear that x=s is a root, hence if we find one additional real root we are done as imaginary roots must exist in pairs.
Here is where I deviated against the solutions. I plugged in x=a,b,c respectively and considered all the cases separated by magnitude and showed that there must exist a pair to the left/right of s such that it is positive and negative, hence there exists another real root.