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

Here,

After simplifying the equation

x² - (a + b)x ab + x² - (b + c)x + bc + x² - (c + a)x + ca

3x² - 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 = b² - 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(a² + b² + c² + 2ab + 2bc+ 2ca) - 12(ab + bc + ca)

= 4(a² + b² + c² + 2ab + 2bc+ 2ca) - 12ab - 12bc - 12ca

= 4(a² + b² + c² + 2ab + 2bc+ 2ca - 3ab - 3bc - 3ca)

= 4(a² + b² + c² - ab - bc - ca)

 

D = 4(a² + b² + c² - ab - bc - ca)

= 2[2a² + 2b² + 2c² - 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(a² + b² + c² - ab - bc - ca)

= 4(a² + b² + c² - aa - bb - cc)

= 4(a² + b² + c² - a² - b² - c²)

= 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