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

Sakanksha Deo by Sakanksha Deo , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

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

2 Helpful 0 Interesting 0 Brilliant 0 Confused
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

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...