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Sample Problem On Quadratic Equation.

Tapas Kumar Maity
19/05/2017 0 0

If 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. 

 

 

 

 

 

 

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