Question

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

Answer 1

If roots of eq are equal

Then , D = 0

b^2 - 4ac = 0

( b - c )^2 - 4 ( a - b ) ( c - a ) = 0

b^2+ c^2 -2bc -4(ac - a^2 + ab - bc) =0

b^2 + c^2 - 2bc + 4bc + 4a^2 - 4ab - 4ac = 0

b^2 + c^2 + 4a^2 + 2bc - 4ac - 4ab = 0

( b + c - 2a ) ^2 = 0

b + c - 2a = 0

b + c = 2a

Therefore , it is in AP

Answer 2

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

 

Roots are equal. So discriminant is zero.    

 

Therefore,  

 

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

 

As 2a = b + c, they can be the three consecutive terms of an AP, because,

In any AP, if three consecutive terms are taken, then the sum of first and third terms is twice the second term.

Hence proved!!!