Question

The roots of the equation $f(x)=a(x-b)(x-c)+b(x-c)(x-a)+c(x-a)(x-b)=0$ ( $a,b,c$ are distinct and real ) are always :

• A. positive
• B. negative
• C. real
• D. unreal

Answer 1

Answer: (C)

$a\left(x^2-(b+c)x+bc\right)+b\left(x^2-(c+a)x+ac\right)+c\left(x^2-(a+b)x+ab\right)=0$
$(a+b+c)x^2-2x(ab+bc+ca)+3abc=0$
$D=4(ab+bc+ca{)}^2-12abc(a+b+c)$
$=4\left[a^2{b}^2+b^2{c}^2+c^2{a}^2+2abc(a+b+c)-3abc(a+b+c)\right]$
$=4\left[a^2{b}^2+b^2{c}^2+c^2{a}^2-abc(a+b+c)\right]$
$=2\left[(ab-bc{)}^2+(bc-ca{)}^2+(ca-ab{)}^2\right]>0$