Prove that both the roots of equations(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)

1.	Prove that both the roots of equations(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
Answer» The given equation is; (x - a)(x - b)+(x - b)(x - c)+(x - c)(x - a) =0\xa0{tex}\\Rightarrow{/tex}\xa0(x2\xa0- ax - bx + ab) + (x2- bx - cx + bc) +(x2\xa0- cx - ax + ac) = 0{tex}\\Rightarrow{/tex}3x2\xa0- 2x(a + b + c) + (ab + bc + ca) = 0.....(1).\xa0Discriminant \'D\' of quadratic equation (1) is given by;{tex}\\therefore{/tex}\xa0D = 4(a + b + c)2\xa0- 12(ab + bc\xa0+ ca)= 4[(a + b + c)2\xa0- 3(ab + bc + ca)]= 4(a2 + b2 + c2\xa0- ab - bc\xa0- ca)= 2(2a2\xa0+ 2b2 + 2c2 -\xa02ab - 2bc\xa0- 2ca)= 2 [(a - b)2 + (b - c)2 + (c - a)2] ≥ 0[{tex}\\because{/tex}\xa0(a - b)2\xa0≥ 0, (b - c)2\xa0{tex}{/tex}\xa0≥ 0 and (c - a)2\xa0{tex}{/tex}\xa0≥ 0].This shows that both the roots of the given equation are real.For equal roots, we must have D = 0.Now, D = 0 {tex}\\Rightarrow{/tex}\xa0(a - b)2\xa0+ (b - c)2\xa0+ (c - a)2= 0{tex}\\Rightarrow{/tex}\xa0(a - b) = 0, (b - c) = 0 and (c - a) = 0 (sum of squares can be zero only if they all are equal to 0){tex}\\Rightarrow{/tex}\xa0a = b = c.Hence, the roots are equal only when a = b = c

Prove that both the roots of equations(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0