#### 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.

#### Solution

The quadric equation is (x - a)(x - b) +(x - b)(x - c)+ (x - c)(x - a) = 0

Here,

After simplifying the equation

x^{2} - (a + b)x ab + x^{2} - (b + c)x + bc + x^{2} - (c + a)x + ca

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

a = 3, b = - 2(a + b + c) and c = (ab + bc + ca)

As we know that D = b^{2} - 4ac

Putting the value of a = 3, b = - 2(a + b + c) and c = (ab + bc + ca)

D = {- 2(a + b + c)}^{2} - 4 x (3) x (ab + bc + ca)

= 4(a^{2} + b^{2} + c^{2} + 2ab + 2bc+ 2ca) - 12(ab + bc + ca)

= 4(a^{2} + b^{2} + c^{2} + 2ab + 2bc+ 2ca) - 12ab - 12bc - 12ca

= 4(a^{2} + b^{2} + c^{2} + 2ab + 2bc+ 2ca - 3ab - 3bc - 3ca)

= 4(a^{2} + b^{2} + c^{2} - ab - bc - ca)

D = 4(a^{2} + b^{2} + c^{2} - ab - bc - ca)

= 2[2a^{2} + 2b^{2} + 2c^{2} - 2ab - 2ac - 2bc]

= 2[(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]

Since, D > 0. So the solutions are real

Let a = b = c

Then

D = 4(a^{2} + b^{2} + c^{2} - ab - bc - ca)

= 4(a^{2} + b^{2} + c^{2} - aa - bb - cc)

= 4(a^{2} + b^{2} + c^{2} - a^{2} - b^{2} - c^{2})

= 4 x 0

Thus, the value of D = 0

Therefore, the roots of the given equation are real and but they are equal only when, a = b = c

Hence proved