Both the roots of the equation (x – b) (x – c) + (x – a) (x – c) + (x – a) (x – b) = 0 are always

1) positive

2) real

3) negative

4) none of these

Solution: (2) real

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

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

Discriminant = 4 (a + b + c)2 – 12 (ab + bc + ca)

= 4 (a2 + b2 + c2 – ab – bc – ca)

= 2{(a – b)2 + (b – c)2 + (c – a)2} which is always positive.

Both the roots are real.

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