Question

if the sum of the roots of the quadratic equation ax^2+bx+c=0 is equal to the sum of the squares of their reciprocals.Show that bc^2,ca^2,ab^2 are in A.P.​

Answer 1

Given that

P(x) = ax²+bx+c = 0

Now,

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


As we know that :-

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


Putting down the values in the first part :-

$\alpha + \beta = {( \frac{ \alpha + \beta }{ \alpha \beta } )}^{2} - \frac{2}{ \alpha \beta } \\ \\ \\ \frac{ - b}{a} = {(\frac{ \frac{ - b}{a} }{ \frac{c}{a} } )}^{2} - \frac{2}{ \frac{c}{a} } \\ \\ \\ \frac{ - b}{a} = \frac{ {b}^{2} }{ {c}^{2} } - \frac{2a}{c} \\ \\ \\ \frac{ - b}{a} = \frac{ {b}^{2} - 2ac }{ {c}^{2} } \\ \\ \\ - b {c}^{2} = a {b}^{2} - 2c {a}^{2} \\ \\ \\ 2c {a}^{2} = a {b}^{2} + b {c}^{2}$


Now,

If A, B, C are in AP
Then,

2B= A+C


Similarly in the above case :-

2ca²=ab²+bc²

Thus,

BC², CA², AB² are in AP