# nature of roots

• Jan 15th 2012, 02:12 AM
arangu1508
nature of roots
Find the nature of roots for
(x-a)(x-b) + (x-c)(x-a) + (x-b)(x-c) = 0
I started solving the above equation.
This was deduced to
3x2 – 2(a + b+c)x +(ab+bc+ca) =0
from this, the discreminant
4(a + b+c)2 -4(3)( ab+bc+ca)
4[a2+b2+c2+2(ab+bc+ca)-3(ab+bc+ca)]
4[a2+b2+c2-(ab+bc+ca)]
From this I am not able to say what type are the roots.
Kindly guide me.
Thanks
• Jan 15th 2012, 03:58 AM
FernandoRevilla
Re: nature of roots
Completing squares we can write the discriminant $D$ in the form

$D=\ldots=(2a-b-c)^2+3(b-c)^2$

This implies $D\geq 0$ for all $a,b,c\in\mathbb{R}$ i.e. all the roots are real. Besides, the roots are equal if and only if $(2a-b-c=0)\;\wedge\; (b-c=0)$ or equivalently if and only if $a=b=c$ .
• Jan 15th 2012, 07:32 AM
arangu1508
Re: nature of roots
Thanks Mr.Fernando Revilla.