Question

a (b-c)x^2+b (c-a)x+c (a-b) has equal roots then prove 1/a+1/c=2/b

Answer 1

Heya mate
The answer is here





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 , it must be

b = 2ac /(a+ c )





hope it helps you

Answer 2

Please see attached images