SOLUTION: Show that the equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c)=0 has equal roots when a=b=c

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Question 153004: Show that the equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c)=0 has equal roots when a=b=c
Answer by jim_thompson5910(28696) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-a%29%28x-b%29%2B%28x-b%29%28x-c%29%2B%28x-a%29%28x-c%29=0 Start with the given equation.


%28x-a%29%28x-a%29%2B%28x-a%29%28x-a%29%2B%28x-a%29%28x-a%29=0 Plug in b=a and c=a


x%5E2-2ax%2Ba%5E2%2Bx%5E2-2ax%2Ba%5E2%2Bx%5E2-2ax%2Ba%5E2=0 FOIL


3x%5E2-6ax%2B3a%5E2=0 Combine like terms.


3%28x-a%29%5E2=0 Factor the left side


%28x-a%29%5E2=0 Divide both sides by 3.


x-a=0 Take the square root of both sides.


x=a Add "a" to both sides.


So the expression %28x-a%29%28x-b%29%2B%28x-b%29%28x-c%29%2B%28x-a%29%28x-c%29 has a root of x=a (with a multiplicity of 2) if a=b=c