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.

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> 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.      Log On


   

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) About Me  (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 . This can be rewritten as . 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.