Question
Prove that the roots of $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are always real and they will be equal if and only if $a=b=c$.
Step 1
This can be written as: \[x^2 - (a+b)x + ab + x^2 - (b+c)x + bc + x^2 - (c+a)x + ac = 0\] This simplifies to: \[3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0\] Show more…
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