Question

If $\alpha $ , $\beta $ are 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) $ac{x^2} + \left( {a + c} \right)cx + {\left( {a + c} \right)^2} = 0$
D) None of the above.

Answer 1

Hint:
We can find the sum and product of the roots $\alpha $, $\beta $ in terms of the coefficient of the given equation. Then we can find the sum and product of the new roots and simplify it further by making substitutions. Then we can find the required equation by substituting the sum and product as coefficients and then by simplification.

Complete step by step solution:
We are given that $\alpha $ , $\beta $ are roots of the equation $a{x^2} + bx + c = 0$ .
We know that the sum of the roots of the quadratic equation of the form $a{x^2} + bx + c = 0$ is given by $\dfrac{{ - b}}{a}$.
 $ \Rightarrow \alpha + \beta = \dfrac{{ - b}}{a}$ … (1)
We know that the product of the roots of the quadratic equation of the form $a{x^2} + bx + c = 0$ is given by $\dfrac{c}{a}$.
 $ \Rightarrow \alpha \beta = \dfrac{c}{a}$ … (2)
Now we need to find the equation whose roots are $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$.
The sum of the roots is given by,
 $ \Rightarrow S = \alpha + \dfrac{1}{\beta } + \beta + \dfrac{1}{\alpha }$
 $ \Rightarrow S = \alpha + \beta + \dfrac{1}{\alpha } + \dfrac{1}{\beta }$
We can take the LCM of the last 2 terms,
 $ \Rightarrow S = \alpha + \beta + \dfrac{{\alpha + \beta }}{{\alpha \beta }}$
On further simplification, we get,
 $ \Rightarrow S = \left( {\alpha + \beta } \right)\left( {1 + \dfrac{1}{{\alpha \beta }}} \right)$
On substituting equations (1) and (2), we get,
 $ \Rightarrow S = \left( {\dfrac{{ - b}}{a}} \right)\left( {1 + \dfrac{1}{{\dfrac{c}{a}}}} \right)$
On simplification we get,
 $ \Rightarrow S = \left( {\dfrac{{ - b}}{a}} \right)\left( {1 + \dfrac{a}{c}} \right)$
On taking LCM we get,
 $ \Rightarrow S = \left( {\dfrac{{ - b}}{a}} \right)\left( {\dfrac{{c + a}}{c}} \right)$
On simplification we get,
 $ \Rightarrow S = \dfrac{{ - b\left( {c + a} \right)}}{{ac}}$
Now the product is given by,
 $ \Rightarrow P = \left( {\alpha + \dfrac{1}{\beta }} \right)\left( {\beta + \dfrac{1}{\alpha }} \right)$
On expanding the bracket, we get,
 \[ \Rightarrow P = \alpha \beta + \alpha \dfrac{1}{\alpha } + \dfrac{1}{\beta }\beta + \dfrac{1}{\beta } \times \dfrac{1}{\alpha }\]
On simplification, we get,
 $ \Rightarrow P = \alpha \beta + 1 + 1 + \dfrac{1}{{\alpha \beta }}$
We can take the LCM, so we will get,
 \[ \Rightarrow P = \dfrac{{{{\left( {\alpha \beta } \right)}^2} + 2\left( {\alpha \beta } \right) + 1}}{{\alpha \beta }}\]
The numerator is the expansion of \[{\left( {\alpha \beta + 1} \right)^2} = {\left( {\alpha \beta } \right)^2} + 2\left( {\alpha \beta } \right) + 1\] . So, we get,
 \[ \Rightarrow P = \dfrac{{{{\left( {\alpha \beta + 1} \right)}^2}}}{{\alpha \beta }}\]
Now we can substitute, equation (2)
 \[ \Rightarrow P = \dfrac{{{{\left( {\dfrac{c}{a} + 1} \right)}^2}}}{{\dfrac{c}{a}}}\]
On taking LCM of numerator we get,
 \[ \Rightarrow P = \dfrac{{{{\left( {\dfrac{{c + a}}{a}} \right)}^2}}}{{\dfrac{c}{a}}}\]
On simplification, we get,
 \[ \Rightarrow P = \dfrac{{a{{\left( {c + a} \right)}^2}}}{{{a^2}c}}\]
On cancelling the common terms, we get,
 \[ \Rightarrow P = \dfrac{{{{\left( {c + a} \right)}^2}}}{{ac}}\]
Now we have the sum and product of the equation. Then the equation is given by,
 ${x^2} - Sx + P = 0$
On substituting the product and sum, we get,
 $ \Rightarrow {x^2} - \dfrac{{ - b\left( {c + a} \right)}}{{ac}}x + \dfrac{{{{\left( {c + a} \right)}^2}}}{{ac}} = 0$
On multiplying throughout with ac, we get,
 $ \Rightarrow ac{x^2} + b\left( {c + a} \right)x + {\left( {c + a} \right)^2} = 0$
Therefore, the required equation is $ac{x^2} + \left( {a + c} \right)bx + {\left( {a + c} \right)^2} = 0$

So, the correct answer is option A.

Note:
We must find the product and sum of the new roots and make them in terms of a, b and c, which are the coefficient of the given equation. The quadratic equation with sum of the roots S and product of the roots P is given by, ${x^2} - Sx + P = 0$. While substituting the sum and product, we must take care that an extra negative sign must be added to the negative sign in the sum of the roots. We must make sure that the denominator of both the sum and product of the roots are the same so that we can cancel the denominator by multiplying throughout the equation with the denominator.