Question

Prove that both the roots of the 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 .

Answer 1

The both roots of the 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

Step 1:

(x-a) (x-b) +(x-b) (x-c) +(x-c) (x-a) =0

Step 2:

x^2 - ax - bx+ab +x^2 - bx - cx + bc +x^2 - cx - ax +ca=0

Step 3:

3x^2 - 2x(a+b+c)+ab+bc+ca=0

Step 4:

Ax^2+Bx+C=0

Step 5:

Comparing the equation in step 3and 4 we get

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

Step 6:

So the roots of the equation will be real and equal only when

B^2-4AC=0.

Step 7:

Putting the value of A, B, C in the equation in step 6 we'll get

(2a+2b+2c)^2 - 4*3(ab+bc+ca)=0

Step 8:

Solving the equation in step 7 we get

4a^2+4b^2+4c^2+8ab+8bc+8ca - 12ab-12bc-12ca=0

2(2a^2+2b^2+2c^2-2ab-2bc-2ca)=0

(a^2 -2ab+b^2) +(b^2 - 2bc+c^2) +(c^2 - 2ca+a^2)=0

(a-b) ^2+(b-c)^2+(c-a)^2=0

Step 9:

As we know that the sum of two or more real squared quantities is equal to zero then the value of each quantity will be zero. Otherwise it's not possible.

Step 10:

a-b=0 b-c=0 c-a=0

a=b b=c c=a

Therefore a=b=c [proved].