document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #530442 by richard1234(7193)\"\" \"About 
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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.\r
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\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "
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