Question

if the roots of equation (a-b)X^2+(b-c)X+(c-a)=0 are equal roots prove that 2a=b+c

Answer 1

For roots to be equal : $b^{2}$ = 4ac;
$(b-c)^{2}$ = 4*(a-b)(c-a)
$b^{2}$ + $c^{2}$ + 2bc = -4$a^{2}$ - 4bc - 4ab  + 4ac
4$a^{2}$ + $b^{2}$ + $c^{2}$ + 4ab - 4ac + 2bc = 0
above equation is similar to $(a+b+c)^{2}$ = 0 , except 'a' replaced by '-2a'
$(-2a + b + c)^{2}$ = 0
-2a + b +c = 0
2a = b + c

Answer 2

$p(x) = (a-b)x^2+(b-c)x+(c-a)=0 \\ \\ \\ A=(a-b) \\ \\ B=(b-c) \\ \\ C=(c-a)$

 

Here given that the roots of p(x) are equal. So the discriminant will be zero.    

 

So,  

 

$B^2-4AC = 0 \\ \\ (b-c)^2-(4(a-b)(c-a)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ac-a^2-bc+ab)) = 0 \\ \\ (b^2-2bc+c^2)-(4(ab-bc+ac-a^2)) = 0 \\ \\ (b^2-2bc+c^2)-(4ab-4bc+4ac-4a^2)=0 \\ \\ b^2-2bc+c^2-4ab+4bc-4ac+4a^2=0 \\ \\ 4a^2+b^2+c^2-4ab+2bc-4ac=0 \\ \\ (2a-b-c)^2=0 \\ \\ 2a-b-c=0 \\ \\ 2a=b+c$

 

Hence proved!!!