Question

If $\alpha {\text{ and }}\beta $ are the roots of the equation $l{x^2} + mx + n = 0,$ then the equation whose roots are ${\alpha ^3}\beta {\text{ and }}\alpha {\beta ^3}{\text{ , is ;}}$
$\left( 1 \right){l^4}{x^2} - nl\left( {{m^2} - 2nl} \right)x + {n^4} = 0$
$\left( 2 \right){l^4}{x^2} + nl\left( {{m^2} - 2nl} \right)x + {n^4} = 0$
$\left( 3 \right){l^4}{x^2} + nl\left( {{m^2} - 2nl} \right)x - {n^4} = 0$
$\left( 4 \right){l^4}{x^2} - nl\left( {{m^2} + 2nl} \right)x + {n^4} = 0$

Answer 1

Hint: The general or standard form of a quadratic equation is given by ; $a{x^2} + bx + c = 0$ where $a,b,c$ are real numbers and $a \ne 0$ ( otherwise it will become a linear equation of degree $1$ ) . Here,$x$ is an unknown variable . The coefficient of ${x^2}$ is $a$ , the coefficient of $x$ is $b$ and $c$ is a constant real number. Example: $7{x^2} + 9x + 6 = 0$ where $a = 7,{\text{ b}} = 9,{\text{ }}c = 6$ .

Complete step by step solution:
The given equation is $l{x^2} + mx + n = 0,$
Let $\alpha {\text{ and }}\beta $ are the two roots of the above quadratic equation , then;
$\left( 1 \right)$ The sum of roots of the quadratic equation $ = - \left( {\dfrac{{{\text{Coefficient of }}x{\text{ }}}}{{{\text{Coefficient of }}{x^2}}}} \right) = - \dfrac{m}{l}$
Therefore, $\alpha + \beta = - \dfrac{m}{l}{\text{ }}......\left( 1 \right)$
$\left( 2 \right)$ The product of roots of the quadratic equation $ = \left( {\dfrac{{{\text{Constant term}}}}{{{\text{Coefficient of }}{x^2}}}} \right) = \dfrac{n}{l}$
Therefore, $\alpha \beta = \dfrac{n}{l}{\text{ }}......\left( 2 \right)$
The general form of quadratic equation in terms of sum of roots and product of roots is given by;
$ \Rightarrow {x^2} - \left( {{\text{sum of roots}}} \right)x + {\text{product of roots = 0}}$
According to the given question, the roots are given as ${\alpha ^3}\beta {\text{ and }}\alpha {\beta ^3}{\text{ , }}$and we have to find out the equation for the given roots, hence we will calculate sum of roots and product of roots;
$\left( 1 \right){\text{Sum of roots = }}{\alpha ^3}\beta + \alpha {\beta ^3}$
$ \Rightarrow {\alpha ^3}\beta + \alpha {\beta ^3} = \alpha \beta \left( {{\alpha ^2} + {\beta ^2}} \right)$
To calculate the value of ${\alpha ^2} + {\beta ^2}$ , we can use the following algebraic identity;
$ \Rightarrow {a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$
Using the above trigonometric identity in the above equation, we get;
$ \Rightarrow {\alpha ^3}\beta + \alpha {\beta ^3} = \alpha \beta \left( {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right)$
We know the values of $\alpha + \beta = - \dfrac{m}{l}$ and $\alpha \beta = \dfrac{n}{l}{\text{ }}$ from equation $\left( 1 \right){\text{ and equation }}\left( 2 \right)$ , put the values in the above equation we get;
$ \Rightarrow {\alpha ^3}\beta + \alpha {\beta ^3} = \dfrac{n}{l}\left( {{{\left( {\dfrac{{ - m}}{l}} \right)}^2} - 2\dfrac{n}{l}} \right)$
Further simplifying the above equation, we get;
$ \Rightarrow {\alpha ^3}\beta + \alpha {\beta ^3} = \dfrac{n}{l}\left( {\dfrac{{{m^2}}}{{{l^2}}} - 2\dfrac{n}{l}} \right){\text{ }}......\left( 3 \right)$
$\left( 2 \right){\text{Product of roots = }}{\alpha ^3}\beta \times \alpha {\beta ^3} = {\left( {\alpha \beta } \right)^4}$
Put the value of $\alpha \beta $ from equation $\left( 2 \right)$, we get;
$ \Rightarrow {\alpha ^3}\beta \times \alpha {\beta ^3} = {\left( {\dfrac{n}{l}} \right)^4}{\text{ }}......\left( 4 \right)$
Now put the values of sum of roots and product of roots from equation $\left( 3 \right){\text{ and }}\left( 4 \right)$ in the general equation, we get the required equation for roots ${\alpha ^3}\beta {\text{ and }}\alpha {\beta ^3}{\text{ , as ;}}$
$ \Rightarrow {x^2} - \dfrac{n}{l}\left( {\dfrac{{{m^2}}}{{{l^2}}} - \dfrac{{2n}}{l}} \right)x + {\left( {\dfrac{n}{l}} \right)^4}{\text{ = 0}}$
Further simplifying the above equation ;
$ \Rightarrow {x^2} - \dfrac{n}{l}\left( {\dfrac{{{m^2} - 2nl}}{{{l^2}}}} \right)x + \dfrac{{{n^4}}}{{{l^4}}}{\text{ = 0}}$
$ \Rightarrow {x^2} - \left( {\dfrac{{n\left( {{m^2} - 2nl} \right)}}{{{l^3}}}} \right)x + \dfrac{{{n^4}}}{{{l^4}}}{\text{ = 0}}$
Taking L.C.M. of the above equation , we get;
$ \Rightarrow \dfrac{{{l^4}{x^2} - nl\left( {{m^2} - 2nl} \right)x + {n^4}{\text{ }}}}{{{l^4}}}{\text{ = 0}}$
Therefore, we get the required equation for the given roots as;
$ \Rightarrow {l^4}{x^2} - nl\left( {{m^2} - 2nl} \right)x + {n^4} = 0$
Hence the correct answer for this question is option $\left( 1 \right)$ .

Note:
An equation of degree $2$ is called a quadratic equation . The degree of a quadratic equation gives the number of roots or zeros, hence the number of roots of a quadratic equation are $2$ . The
roots are calculated by the quadratic equation formula; $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . The term ${b^2} - 4ac$ is called the determinant of the quadratic equation as it determines the type of roots , like : $\left( 1 \right){\text{If }}{b^2} - 4ac \succ 0$ or positive and there will be two distinct real values of $x$ . $\left( 2 \right){\text{If }}{b^2} - 4ac = 0$ then there will be two repeated real number solutions for $x$ . $\left( 3 \right){\text{If }}{b^2} - 4ac \prec 0$ or negative then there we will get complex values for $x$ .