Question

If the sum of roots of the equation $a{x^2} + bx + c = 0$ is equal to the sum of the squares of their reciprocals, then prove that $2{a^2}c = {c^2}b + {b^2}a$.

Answer 1

Hint: let $\alpha $and $\beta $ be the two roots of the quadratic equation. Sum of roots of any quadratic equation $a{x^2} + bx + c = 0$ is $\dfrac{{ - b}}{a}$ and the product of roots is $\dfrac{c}{a}$.
$\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$
Using the relation, sum of roots = sum of square of reciprocals, and substituting the values of $\alpha + \beta $and $\alpha \beta $, we can prove the above expression.

Complete Step by Step Solution:
Given equation: $a{x^2} + bx + c = 0$
Sum of the roots of this equation = $\alpha + \beta = \dfrac{{ - b}}{a}$
Product of roots of this equation = $\alpha \beta = \dfrac{c}{a}$
Given condition: sum of roots = sum of squares of their reciprocals
$ \Rightarrow \alpha + \beta = \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}}$
$ \Rightarrow \alpha + \beta = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{{{(\alpha \beta )}^2}}}............................$(equation $1$)
We have the value of $\alpha + \beta $ and$\alpha \beta $. But we do not have the value of ${\alpha ^2} + {\beta ^2}$. We can manipulate ${\alpha ^2} + {\beta ^2}$in terms of $\alpha + \beta $and $\alpha \beta $.
${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta $
Substituting the value of ${\alpha ^2} + {\beta ^2}$in equation $1$,
$ \Rightarrow \alpha + \beta = \dfrac{{{{(\alpha + \beta )}^2} - 2\alpha \beta }}{{{{(\alpha \beta )}^2}}}......................$(equation $2$)
Now, we know that the sum of roots = $\alpha + \beta = \dfrac{{ - b}}{a}$
And the product of roots = $\alpha \beta = \dfrac{c}{a}$
Substituting the value of $\alpha + \beta $and $\alpha \beta $in equation $2$,
$ \Rightarrow \dfrac{{ - b}}{a} = \dfrac{{{{(\dfrac{{ - b}}{a})}^2} - 2(\dfrac{c}{a})}}{{{{(\dfrac{c}{a})}^2}}}$
$ \Rightarrow \dfrac{{ - b}}{a} = \dfrac{{{b^2} - 2ac}}{{{c^2}}}$
On cross multiplication,
$ \Rightarrow - b{c^2} = a{b^2} - 2{a^2}c$
Rearranging the terms.
$2{a^2}c = a{b^2} + b \to $(proved)

Note:
We proved that $2{a^2}c = {c^2}b + {b^2}a$
Dividing LHS and RHS by $abc$,
$ \Rightarrow \dfrac{{2a}}{b} = \dfrac{c}{a} + \dfrac{b}{c}$
$ \Rightarrow \dfrac{c}{a},\dfrac{a}{b},\dfrac{b}{c}$ are in arithmetic progression.
And reciprocals are in HP.
Remember the formula of sum of roots and product of roots. Using this, you can solve maximum questions of quadratic equations.