Prove that both the roots of the quadratic equation . . .

Question : Prove that both the roots of the quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are real but they are equal only when a=b=c.

Doubt by Afifa

Solution :

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
x(x-b)-a(x-b)+x(x-c)-b(x-c)+x(x-a)-c(x-a)=0
x²-bx-ax+ab+x²-cx-bx+bc+x²-ax-cx+ca=0

3x²-2ax-2bx-2cx-2dx+ab+bc+ca=0

3x²-2(a+b+c)x+(ab+bc+ca)=0

Now, comparing the coefficients

A = 3
B = -2(a+b+c)
C = (ab+bc+ca)

D = B²-4AC
D = [-2(a+b+c)]²-4(3)(ab+bc+ca)

D = 4(a+b+c)²-12(ab+bc+ca)
D = 4(a²+b²+c²+2ab+2bc+2ca)-12(ab-bc-ca)

D = 4a²+4b²+4c²+8ab+8bc+8ca-12ab-12bc-12ca

D = 4a²+4b²+4c²-4ab-4bc-4ca

D = 2[2a²+2b²+2c²-2ab-2bc-2ca]
D = 2[a²+a²+b²+b²+c²+c²-2ab-2bc-2ca]

D = 2[(a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca)]
D = 2[(a-b)²+(b-c)²+(c-a)²]
Clearly
D>0

Hence, the roots are real.

But if a=b=c
then
D = 2[(0)²+(0)²+(0)²]
D=2×0
D=0
Hence, the roots are real and equal.

Class 10