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Both the roots of the equation (x-b)(x-c...

Both the roots of the equation `(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0` are always a. positive b. real c. negative d. none of these

A

positive

B

negative

C

real

D

real and equal

Text Solution

Verified by Experts

The correct Answer is:
C
`(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0`
`rArr 3x^(2) - 2 (a + b + c)x + (ab + bc + ca) = 0`
Now `D = 4 (a + b+ c)^(2) - 12(ab + bc + ca)`
`= 4(a^(2) + b^(2) + c^(2) - ab - bc - ca)`
`= 2[(a - b)^(2) + (b - c)^(2) + (c - a)^(2)]`
which is always positive or zero so roots are real
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