0If the roots of the equation a(b-c) x^2+b(c-a)x+c(a-b)=0 are equal then prove that 1/a + 1/c = 2 /b
Solution :
Since roots of the equation are equal, value of discrinant is zero.
Which give {b(c-a)} ^2 - 4a(b-c).c(a-b)=0
or, b^2c^2 + b^2a^2 - 2ab^2c - 4a^2bc + 4ab^2c + 4a^2c^2 - 4abc^2 =0
or, b^2c^2 + b^2a^2 + 4a^2c^2 + 2ab^2c - 4a^2bc - 4abc^2 = 0
or, ( bc + ab - 2ac) ^2 = 0
or, bc + ab =2ac
Now dividing both sides by abc we get the result.
