Prove that roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always real and can be equal unless a=b=c

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Ioriannis4v · Accepted Answer

3      x    2    −  2  x  (  a  +  b  +  c  )  +  (  a  b  +  b  c  +  c  a  )  =  0    D  =  4  (  a  +  b  +  c      )    2    −  12  (  a  b  +  b  c  +  c  a  )    =  4  [  (  a  +  b  +  c      )    2    −  3  (  a  b  +  b  c  +  c  a  )  ]    =  2  [  2      a    2    +  2      b    2    +  2      c    2    −  2  a  b  −  2  b  c  −  2  c  a  ]    =  2  [  (  a  −  b      )    2    +  (  b  −  c      )    2    +  (  c  −  a      )    2    ]    I  f     D  =  0  ,     t  h  e  n    (  a  −  b      )    2    +  (  b  −  c      )    2    +  (  c  −  a      )    2    =  0    ⇒  a  −  b  =  0  ;  b  =  c  ;  c  =  a    a  =  b  =  c

Prove that roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always real and can be equal unless a=b=c

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