Question

If the roots of equation: a (b – c) x2 + b (c – a) x + c (a – b) = 0 are equal then. Show that 1/a+1/c=2/b

Answer 1

Given quadratic equation,
a(b-c)x^2+b(c-a)^2+c(a-b)=0
also zeroes of given equation are equal.
Therefore,
discriminant=0
b^2-4ac=0

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

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

b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0

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

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

b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0
We know that,
a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2
By above identity we get,
{bc+ab-2ac}^2=0

bc+ab-2ac=0

b(a+c)=2ac

b=2ac/(a+c)

Hence if zeroes of given quadratic equation are equal then,
b=2ac/(a+c)

1/b = (a+c)/2ac

2/b = a/(ac) + c/ac

2/b = 1/c + 1/a

"1/a + 1/c = 2/b"__________proved


I think my answer is capable to clear your confusion..

Answer 2

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