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Question: Both the roots of given equation \((x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0\) are alway...

Both the roots of given equation

(x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 are always

A

Positive

B

Negative

C

Real

D

Imaginary

Answer

Real

Explanation

Solution

Given equation

(x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 can be re-written as 3x2−2(a+b+c)x+(ab+bc+ca)=03x^{2} - 2(a + b + c)x + (ab + bc + ca) = 0

D=4[(a+b+c)2−3(ab+bc+ca)]=4[a2+b2+c2−ab−bc−ac]=2[(a−b)2+(b−c)2+(c−a)2]≥0D = 4\lbrack(a + b + c)^{2} - 3(ab + bc + ca)\rbrack = 4\lbrack a^{2} + b^{2} + c^{2} - ab - bc - ac\rbrack = 2\lbrack(a - b)^{2} + (b - c)^{2} + (c - a)^{2}\rbrack \geq 0

Hence both roots are always real.