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# 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. - Mathematics

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#### 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

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

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#### APPEARS IN

RD Sharma Solution for Class 10 Maths (2018 (Latest))