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Hint: First we will find the sum and product of the roots of the given equation using the formula of sum and product of the roots of the quadratic equation. Then will find the sum and product of the given roots. Finally, I will use ${x^2} - sx + p = 0$ where $s$ is the sum of roots and $p$ is the product of roots.
Formula Used: Let the quadratic equation ${a^2}x + bx + c = 0$ then,
Sum of roots $\dfrac{{ - b}}{a}$
And Product of roots $\dfrac{c}{a}$
Complete step by step solution: Given, equation ${a^2}x + bx + c = 0$
Sum of roots $\dfrac{{ - b}}{a}$
$\alpha + \beta = \dfrac{{ - b}}{a}$
Product of roots $\dfrac{c}{a}$
$\alpha \beta = \dfrac{c}{a}$
Sum of roots of equation have roots $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$
$\alpha + \dfrac{1}{\beta } + \beta + \dfrac{1}{\alpha } = \left( {\alpha + \beta } \right) + \left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)$
After simplification, we get
$ = \left( {\alpha + \beta } \right) + \left( {\dfrac{{\alpha + \beta }}{{\alpha \beta }}} \right)$
$ = \left( {\dfrac{{ - b}}{a}} \right) + \left( {\dfrac{{\dfrac{{ - b}}{a}}}{{\dfrac{c}{a}}}} \right)$
After solving, we get
$ = \left( {\dfrac{{ - b}}{a}} \right) + \left( {\dfrac{{ - b}}{c}} \right)$
$ = \dfrac{{ - bc - ba}}{{ac}}$
$ = \dfrac{{ - b\left( {a + c} \right)}}{{ac}}$
Products of roots $ = \left( {\alpha + \dfrac{1}{\beta }} \right)\left( {\beta + \dfrac{1}{\alpha }} \right)$
$ = \alpha \beta + 1 + 1 + \dfrac{1}{{\alpha \beta }}$
After solving we get
$ = \dfrac{c}{a} + 2 + \dfrac{a}{c}$
$ = \dfrac{{\left( {{c^2} + 2ac + {a^2}} \right)}}{{ac}}$
$ = \dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}}$
We know if s is the sum of roots and p is product then the quadratic equation is ${x^2} - sx + p = 0$.
Then, the equation is ${x^2} + \dfrac{{b\left( {a + c} \right)}}{{ac}}x + \dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}} = 0$
$ac{x^2} + b\left( {a + c} \right)x + {\left( {a + c} \right)^2} = 0$
Hence, option A is correct.
Note: For easy calculations students should find the sum and product of roots of the required equation and to avoid any complicated calculations. Then, use ${x^2} - sx + p = 0$ to find the required quadratic equation.
Formula Used: Let the quadratic equation ${a^2}x + bx + c = 0$ then,
Sum of roots $\dfrac{{ - b}}{a}$
And Product of roots $\dfrac{c}{a}$
Complete step by step solution: Given, equation ${a^2}x + bx + c = 0$
Sum of roots $\dfrac{{ - b}}{a}$
$\alpha + \beta = \dfrac{{ - b}}{a}$
Product of roots $\dfrac{c}{a}$
$\alpha \beta = \dfrac{c}{a}$
Sum of roots of equation have roots $\alpha + \dfrac{1}{\beta }$ and $\beta + \dfrac{1}{\alpha }$
$\alpha + \dfrac{1}{\beta } + \beta + \dfrac{1}{\alpha } = \left( {\alpha + \beta } \right) + \left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)$
After simplification, we get
$ = \left( {\alpha + \beta } \right) + \left( {\dfrac{{\alpha + \beta }}{{\alpha \beta }}} \right)$
$ = \left( {\dfrac{{ - b}}{a}} \right) + \left( {\dfrac{{\dfrac{{ - b}}{a}}}{{\dfrac{c}{a}}}} \right)$
After solving, we get
$ = \left( {\dfrac{{ - b}}{a}} \right) + \left( {\dfrac{{ - b}}{c}} \right)$
$ = \dfrac{{ - bc - ba}}{{ac}}$
$ = \dfrac{{ - b\left( {a + c} \right)}}{{ac}}$
Products of roots $ = \left( {\alpha + \dfrac{1}{\beta }} \right)\left( {\beta + \dfrac{1}{\alpha }} \right)$
$ = \alpha \beta + 1 + 1 + \dfrac{1}{{\alpha \beta }}$
After solving we get
$ = \dfrac{c}{a} + 2 + \dfrac{a}{c}$
$ = \dfrac{{\left( {{c^2} + 2ac + {a^2}} \right)}}{{ac}}$
$ = \dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}}$
We know if s is the sum of roots and p is product then the quadratic equation is ${x^2} - sx + p = 0$.
Then, the equation is ${x^2} + \dfrac{{b\left( {a + c} \right)}}{{ac}}x + \dfrac{{{{\left( {a + c} \right)}^2}}}{{ac}} = 0$
$ac{x^2} + b\left( {a + c} \right)x + {\left( {a + c} \right)^2} = 0$
Hence, option A is correct.
Note: For easy calculations students should find the sum and product of roots of the required equation and to avoid any complicated calculations. Then, use ${x^2} - sx + p = 0$ to find the required quadratic equation.
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