Question

if the sum of the roots of the quadratic equation ax2+bx+c=0 is equal to the sum of the squares of their reciprocals then prove that 2a2=c2b+b2a

Answer 1

 

Let m and n be the roots of the given quadratic equation. 

Sum of roots 
= m+n 
= -b/a 
Product of roots 
= mn 
= c/a 

Sum of squares of reciprocals of the roots 
= 1/m^2 + 1/n^2 
= (m^2 + n^2)/(mn)^2 
= {(m+n)^2 - 2mn}/(mn)^2 
={(-b/a)^2 - 2c/a}/(c/a)^2 
=(b^2 - 2ac)/c^2 

Now, m+n = 1/m^2 + 1/n^2 
or -b/a = (b^2 - 2ac)/c^2 
or b^2/c^2 + b/a = 2a/c 
or b^2/ac + bc/a^2 = 2.