Question

35. If the roots of the Quadratic equation (a-b) x2 + (b_c)x+ (c-a) = 0 are equal, then
prove that 2a=b+c.
OR
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Answer 1

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

 

Roots are equal. So discriminant is zero.    

 

Therefore,  

 

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

 

Hence proved!