Question

If $\alpha $,$\beta $ are the roots of the equation $a{x^2} + bx + c = 0$ then the equation whose roots are $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$ is
a) $ac{x^2} + \left( {a + c} \right)bx + {\left( {a + c} \right)^2} = 0$
b) $ab{x^2} + \left( {a + c} \right)bx + {\left( {a + c} \right)^2} = 0$
c) $ac{x^2} + \left( {a + b} \right)cx + a{\left( {a + c} \right)^2} = 0$
d) None of the above

Answer 1

Hint: We know that if $\alpha $,$\beta $ are the roots of the equation $a{x^2} + bx + c = 0$ then the sum of roots is given by $\alpha + \beta = - \dfrac{b}{a}$ and product of roots is given by $\alpha \beta = \dfrac{c}{a}$. We will use this information to find the equation whose roots are $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$.
Complete step-by-step answer:
It is given that $\alpha $,$\beta $ are the roots of the equation $a{x^2} + bx + c = 0$. So we can say that sum of roots is given by $\alpha + \beta = - \dfrac{b}{a} \cdots \cdots \left( 1 \right)$ and product of roots is given by $\alpha \beta = \dfrac{c}{a} \cdots \cdots \left( 2 \right)$. In this problem, we have to find an equation whose roots are $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$. First we will find sum of these two roots.
Let us find the sum of roots $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$.
Therefore, we get
$\alpha + \dfrac{1}{\beta } + \beta + \dfrac{1}{\alpha }$
$ = \left( {\alpha + \beta } \right) + \left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)$
$ = \left( {\alpha + \beta } \right) + \left( {\dfrac{{\beta + \alpha }}{{\alpha \beta }}} \right)$
$ = \left( {\alpha + \beta } \right) + \left( {\dfrac{{\alpha + \beta }}{{\alpha \beta }}} \right) \cdots \cdots \left( 3 \right)$
Now from $\left( 1 \right)$ and $\left( 2 \right)$, we will substitute values of $\alpha + \beta $ and $\alpha \beta $ in the equation $\left( 3 \right)$. So, we get
$\alpha + \dfrac{1}{\beta } + \beta + \dfrac{1}{\alpha }$
$ = \left( { - \dfrac{b}{a}} \right) + \left( {\dfrac{{ - \dfrac{b}{a}}}{{\dfrac{c}{a}}}} \right)$
$ = - \dfrac{b}{a} - \dfrac{b}{c}$
$ = - b\left( {\dfrac{1}{a} + \dfrac{1}{c}} \right)$
$ = - b\left( {\dfrac{{c + a}}{{ac}}} \right)$
$ = - b\left( {\dfrac{{a + c}}{{ac}}} \right)$
Therefore, we can say that the sum of roots $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$ is $ - b\left( {\dfrac{{a + c}}{{ac}}} \right)$.
Now we will find the product of roots $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$. Therefore, we get
$\left( {\alpha + \dfrac{1}{\beta }} \right)\left( {\beta + \dfrac{1}{\alpha }} \right)$
$ = \alpha \beta + \left( \alpha \right)\left( {\dfrac{1}{\alpha }} \right) + \left( {\dfrac{1}{\beta }} \right)\beta + \left( {\dfrac{1}{\beta }} \right)\left( {\dfrac{1}{\alpha }} \right)$
$ = \alpha \beta + 1 + 1 + \dfrac{1}{{\beta \alpha }}$
$ = \alpha \beta + 2 + \dfrac{1}{{\alpha \beta }} \cdots \cdots \left( 4 \right)$
Now from $\left( 2 \right)$, we will substitute value of $\alpha \beta $ in the equation $\left( 4 \right)$. So, we get
$\left( {\alpha + \dfrac{1}{\beta }} \right)\left( {\beta + \dfrac{1}{\alpha }} \right)$
$ = \dfrac{c}{a} + 2 + \dfrac{1}{{\dfrac{c}{a}}}$
$ = \dfrac{c}{a} + 2 + \dfrac{a}{c}$
$ = \dfrac{{{c^2} + 2ac + {a^2}}}{{ac}}$
$ = \dfrac{{{a^2} + 2ac + {c^2}}}{{ac}}$
Now we are going to use the formula ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$ in the numerator, we get
$\left( {\alpha + \dfrac{1}{\beta }} \right)\left( {\beta + \dfrac{1}{\alpha }} \right) = \dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}}$
Therefore, we can say that the product of roots $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$ which is $\dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}}$.
If the sum and product of two roots is known then we can say that the quadratic equation will be
${x^2} - $ (sum of roots) $x + $ (product of roots) $ = 0$
$ \Rightarrow {x^2} - \left( {\dfrac{{ - b\left( {a + c} \right)}}{{ac}}} \right)x + \left[ {\dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}}} \right] = 0$
$ \Rightarrow \dfrac{{ac{x^2} + b\left( {a + c} \right)x + {{\left( {a + c} \right)}^2}}}{{ac}} = 0$
$ \Rightarrow ac{x^2} + \left( {a + c} \right)bx + {\left( {a + c} \right)^2} = 0$
Therefore, the required equation is $ac{x^2} + \left( {a + c} \right)bx + {\left( {a + c} \right)^2} = 0$.
Therefore, option (a) is correct.
Note: The standard form of quadratic equation is $a{x^2} + bx + c = 0$ where $a \ne 0$. Every quadratic equation has exactly two roots. For the roots of quadratic equation, there are three possible cases:
$\left( 1 \right)$ Roots are real and distinct $\left( 2 \right)$ Roots are real and equal $\left( 3 \right)$ Roots are imaginary numbers.