Question

If the roots of the equation, (b - c)x 2 + (c - a)x + (a - b) = 0 are equal, prove that 2b = a + c

Answer 1

general form is Ax² + Bx + C= 0
if A + B + C = 0 
then zeroes of polynomial are 1 and C/A
here A + B + C = b -c +c - a + a - b = 0
therefore zeroes are 1 and (a - b) / (b - c)
its given that they are equal
∴ (a - b) / (b - c) = 1
⇒a - b = b - c
2b = a + c
hence proved
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Answer 2

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

 

Roots are equal. So discriminant is zero.    

 

Therefore,  

 

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

 

Hence proved!!!