Question

If the sum of the roots of the equation ax2+bx+c=0 is equal to the sum of their squares of their reciprocals, then prove that 2 a2c=c2b+b

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

The given equation can be solved by

1) Let the roots of equation be m and n.

2) According to the question,

 $m+n = \frac{1}{m^2} +\frac{1}{n^2}$

3) Solving the RHS of the equation

$m+n = \frac{m^2 + n^2}{m^2n^2}$

4) Further solving the RHS

$m+n = \frac{(m+n)^2 - 2mn}{m^2 n^2}$

5) The given equation is $ax^2 + bx+c =0$

6) sum of roots, m+n = -b/a

7) Product of roots, mn = c/a

8) Substituting the values in the equation, gives $(\frac{-b}{a} ) = \frac{(\frac{-b}{a})^2 - 2 \frac{c}{a} }{(\frac{c}{a} )^2}$

9) cross multiplying the two parts, $\frac{b c^2}{a^3} = \frac{b^2- 2ac}{a^2}$

10) On simplifying we will have 2a^2*c = c^2*b + b^2*a