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Question

Both the roots of the equation, (if given that a,b,c are real) (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0 are always :

A
Positive
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B
Negative
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C
Real
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D
None of these
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Solution

The correct option is C Real

Solve using assumption

Let a = 0, b = 1, c = 2. The expression becomes 3x2 -6x + 2 = 0. The discriminant of this exp is positive. You can try with some other values. You will always get real roots.

Alternatively: -

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

x2 - cx - bx + bc + x2 - cx - ax + ac + x2 - bx - ax + ab = 0

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

Discriminant = b2 - 4ac

4(a + b + c)2 - 4 x 3 x (ab + bc + ca)

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

4{a2 + b2 + c2 - (ab + bc + ca)}

The discriminant of this exp is positive. You will always get real roots.


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