Equality

Algebra Level 4

If the roots of the equation ( a b ) x 2 + ( b c ) x + ( c a ) = 0 (a - b)x^{2} + (b - c)x + (c - a) = 0 are equal. Then:

1 b = 1 a + 1 c \frac{1}{b} = \frac{1}{a} + \frac{1}{c} 2b = a + c 2c = a + b 2a = b + c

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2 solutions

Let f ( x ) = ( a b ) x 2 + ( b c ) x + ( c a ) f(x) = (a-b)x^2 + (b-c)x + (c-a) . We see that f ( 1 ) = 0 f(1) = 0 . Therefore, since both roots are equal, f ( x ) = k ( x 1 ) 2 = k x 2 2 k x + k f(x) = k(x-1)^2 = kx^2 - 2kx + k . We see that the coefficient of x 2 x^2 and the constant term are equal.

Therefore, a b = c a 2 a = b + c a-b = c -a \implies 2a = b + c

Pranjal Jain
Mar 10, 2015

Since roots are equal, D=0.

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

Now note that this equation is symmetric in b and c. Therefore, only possible answer which is symmetrical in b and c is 2 a = b + c 2a=b+c

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