Question

If the roots of the quadratic equation (n-a) (n-b) + (n-b) (n-c) + (n-c) (n-a) = 0 are equal. Show that a=b=c

Answer 1

Answer:

Open the brackets and solve.

You get:

n^2 - bn - an + ab + n^2 - cn - bn + bc + n^2 - cn - an + ca = 0

=> 3n^2 - 2n(b + a + c) + ab + bc + ca= 0

Roots of this equation = equal

Thus, b^2 - 4ac = 0

=> (2(b + a + c))^2 - 4(3)(ab + bc + ca) = 0

=> 4(b + a + c)^2 - 4(3)(ab + bc + ca) = 0

=> 4 [ (b + c + a)^2 - 3(ab + bc + ca)] = 0

=> (b + a + c)^2 - 3(ab + bc + ca) = 0

=> b^2 + a^2 + c^2 + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca = 0

=> b^2 + a^2 + c^2 - ab - bc - ca = 0

If we take b = a = c = x :

=> x^2 + x^2 + x^2 - x^2 - x^2 - x^2 = 0

=> 0 = 0

L.H.S. = R.H.S.

thus, this is only possible when a = b = c

hence proved.

Hope it Helps!!