Question

If the roots of the quadratic equation

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

are equal, then show that a = b = c.

Answer 1

Solution:

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

Multiplying the terms we get,

⇒ [ x² - bx - ax + ab ] + [ x² - cx - bx + bc ] + [ x² - ax - cx + ac ] = 0

⇒ [ x² + x² + x² - ax - ax - bx - bx - cx - cx + ab + bc + ac ] = 0

⇒ 3x² - 2 ( a + b + c )x + ( ab + bc + ac ) = 0

Now since the equation is a long one, we can solve it using the quadratic formula.

$\text{Quadratic Formula} : \dfrac{ \text{ - b }\:\pm\:\sqrt{ b^2 - 4ac}}{2a}$

Here,

• 'a' refers to coefficient of x²
• 'b' refers to coefficient of x
• 'c' refers to constant part

Calculation of Discriminant:

Discriminant: b² - 4ac

According to the question,

• a = 3
• b = -2 ( a + b + c )
• c = ab + bc + ac

⇒ Discriminant = ( -2 ( a + b + c ) )² - 4 ( 3 ) ( ab + bc + ac )

⇒ Discriminant = 4 ( a² + b² + c² + 2ab + 2bc + 2ac ) - 12 ( ab + bc + ac )

⇒ Discriminant = 4 ( a² + b² + c² ) + 8ab + 8bc + 8ac - 12ab - 12bc - 12ac

⇒ Discriminant = 4 ( a² + b² + c² ) - 4ab - 4bc - 4ac

⇒ Discriminant = 4 ( a² + b² + c² - ab - bc - ac )

Since the roots are equal, Discriminant is zero.

⇒ 4 ( a² + b² + c² - ab - bc - ac ) = 0

Dividing by 2 on both sides we get,

⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

⇒ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ac + a² ) = 0

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

We know that square terms will be zero only when individual terms are zero.

⇒ a - b = 0 ; b - c = 0 ; a - c = 0

⇒ a = b ; b = c ; a = c

⇒ a = b = c

Hence Proved !!