SOLUTION: show that the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always real and they cannot be equal unless a=b=c.
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Question 878896: show that the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always real and they cannot be equal unless a=b=c.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
The x^2 coefficient is 3. The x coefficient is (-a-b) + (-b-c) + (-c-a) = -2(a+b+c). The constant term is ab + bc + ca.
Then the discriminant is ^2 - 4(3)(ab + bc + ca) = 4(a^2 + b^2 + c^2 - ab - bc - ca))
. This can be rewritten as  = 2((a-b)^2 + (b-c)^2 + (c-a)^2))
. This expression is always non-negative, so the roots are necessarily real. It follows that the roots are equal iff a=b=c, since we want the discriminant equal to 0.
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