Question

if the sum of the roots of the quadratic equation ax square + bx + c is equal to zero is equal to the sum of the squares of the reciprocals then prove that 2 a square c = c square b + b square a

Answer 1

$\alpha + \beta = ({\frac{1}{ \alpha }})^{2} + ( {\frac{1}{ \beta}})^{2}$

$\alpha + \beta = \frac{ { \beta }^{2} + { \alpha}^{2} }{ { \alpha }^{2} { \beta }^{2} }$

$( \alpha + \beta) { \alpha }^{2} { \beta }^{2} = { \alpha }^{2} + { \beta }^{2}$

$( \alpha + \beta ) \times \alpha \beta ( \alpha \beta ) = ( { \alpha + \beta })^{2} - \alpha \beta$

Now ,

$\alpha + \beta = \frac{ - b}{a}$

$\alpha \beta = \frac{c}{a}$

$\frac{ - b}{a} \times \frac{ {c}^{2} }{ {a}^{2} } = \frac{ {b}^{2} }{ {a}^{2} } \times \frac{c}{a}$

$\frac{ - a}{b} = \frac{a}{c}$

$- c = b$

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