Question

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

Answer 1

Hint: Every quadratic equation is of the form \[a{x^2} + bx + c = 0\], here, \[a,b,c\] are real and rational numbers.
We are going to find the roots of the equation or assume the roots of the equation.
From that, we will solve the equation.

Formula used: Sum of the roots:
\[\alpha + \beta = \dfrac{{ - b}}{a} = - \dfrac{{co.(x)}}{{co.({x^2})}}\]Where, \[co.(x) = \]coefficient of\[x\], \[co.({x^2})\]=coefficient of \[{x^2}\]
Product of the roots:
\[\alpha \beta = \dfrac{c}{a} = \dfrac{{cons.}}{{co.({x^2})}}\], where cons. = constant term, \[co.({x^2})\]= coefficient of \[{x^2}\]

Complete step-by-step answer:
It is given \[a{x^2} + bx + c = 0\]
Let \[\alpha \] and \[\beta \] be the two roots of the above quadratic equation.
Now we have to use the formula to get the sum and product of the roots of a quadratic equation.
In this given condition is stated as, the sum of the roots of the quadratic equation is equal to the sum of the squares of their reciprocals.
Sum of roots is \[\alpha + \beta \]
Sum of the squares of their reciprocals is \[\dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}}\]
\[\alpha + \beta = \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}}....\left( 1 \right)\]
Taking LCM as in the RHS of equation$\left( 1 \right)$, we get
\[\alpha + \beta \]=\[\dfrac{{{\beta ^2} + {\alpha ^2}}}{{{\alpha ^2}{\beta ^2}}}\]
Here, we can write \[{\beta ^2} + {\alpha ^2}\]as\[{\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \] by using the formula,
That is \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
We can write it as, \[{\left( {a + b} \right)^2} - 2ab = {a^2} + {b^2}\]
Here, \[a = \alpha \] and \[b = \beta \]
So we can write it as, \[\alpha + \beta = \dfrac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{{\alpha ^2}{\beta ^2}}}....\left( 2 \right)\]
Also, \[{\alpha ^2}{\beta ^2} = {(\alpha \beta )^2}\]
Using the sum of roots property, \[\alpha + \beta \]=\[\dfrac{{ - b}}{a}\]
And product of roots property, \[\alpha \beta = \dfrac{c}{a}\]
Substitute these values in equation\[\left( 2 \right)\]we get,
\[ - \dfrac{b}{a} = \dfrac{{{{\left( { - \dfrac{b}{a}} \right)}^2} - 2\left( {\dfrac{c}{a}} \right)}}{{{{\left( {\dfrac{c}{a}} \right)}^2}}}\]
Square of minus is plus, we get
\[ - \dfrac{b}{a} = \dfrac{{\dfrac{{{b^2}}}{{{a^2}}} - \dfrac{{2c}}{a}}}{{\left( {\dfrac{{{c^2}}}{{{a^2}}}} \right)}}\]
Taking LCM as \[{a^2}\] and we write it as,
\[ - \dfrac{b}{a} = \dfrac{{\dfrac{{{b^2}}}{{{a^2}}} - \dfrac{{2c \times a}}{{a \times a}}}}{{\dfrac{{{c^2}}}{{{a^2}}}}}\]
On multiplying the terms we get,
\[ - \dfrac{b}{a} = \dfrac{{\dfrac{{{b^2}}}{{{a^2}}} - \dfrac{{2ac}}{{{a^2}}}}}{{\dfrac{{{c^2}}}{{{a^2}}}}}\]
Taking LCM as common
\[ - \dfrac{b}{a} = \dfrac{{\dfrac{{{b^2} - 2ac}}{{{a^2}}}}}{{\dfrac{{{c^2}}}{{{a^2}}}}}\]
In the RHS, we can write it as in the form of reciprocal terms, that is
Here, \[\dfrac{{{c^2}}}{{{a^2}}}\]in denominator can be written as multiply of\[\dfrac{{{a^2}}}{{{c^2}}}\] into numerator
That is, \[ - \dfrac{b}{a} = \dfrac{{{a^2}}}{{{c^2}}}\left( {\dfrac{{{b^2} - 2ac}}{{{a^2}}}} \right)\]
Cancelling \[{a^2}\] we get,
\[ - \dfrac{b}{a} = \dfrac{1}{{{c^2}}}({b^2} - 2ac)\]
Let us take cross multiply the denominator from right hand side to left hand side
\[ - b{c^2} = a\left( {{b^2} - 2ac} \right)\]
On multiply the RHS we get,
\[ - b{c^2} = a{b^2} - 2{a^2}c\]
Take \[ - 2{a^2}c\] from right hand side to left hand side
\[2{a^2}c - b{c^2} = a{b^2}\]
Take \[ - b{c^2}\] from left hand side to right hand side
\[2{a^2}c = a{b^2} + b{c^2}\]
Hence proved.

Note: The quadratic equation \[a{x^2} + bx + c = 0\], we can write the equation as like also,
\[{x^2} - (\alpha + \beta )x + \alpha \beta = 0\]
The sum of roots and product of roots occupies the position in the coefficient of \[x\] and coefficient of constant.
Also, note that the coefficient of \[{x^2}\] is one.