Question

For the equation $l{x}^2+mx+n=0,l\ne 0$. If $\alpha \&\beta$ are roots of the equation $m^3+l^{2}n+{\ln }^2=3lmn$ then

• A. $\alpha ={\beta }^2$
• B. ${\alpha }^3=\beta$
• C. $\alpha +\beta =\alpha \beta$
• D. $\alpha \beta =1$

Answer 1

The correct option is A

$\alpha ={\beta }^2$

Explanation for correct option:

Step1.Finding roots of equation,

$l{x}^2+mx+n=0,l\ne 0$

$\alpha \&\beta$ are the are roots of the equation,

$\begin{array}{rcl}\mathit{S}\mathit{u}\mathit{m}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{r}\mathit{o}\mathit{o}\mathit{t}\mathit{s}\mathbf{}& \mathbf{=}& \frac{\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{e}\mathbf{f}\mathbf{f}\mathbf{i}\mathbf{c}\mathbf{i}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{x}}{\mathbf{c}\mathbf{o}\mathbf{e}\mathbf{f}\mathbf{f}\mathbf{i}\mathbf{c}\mathbf{i}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{x}}^{\mathbf{2}}}\\ \alpha +\beta & =& \frac{-m}{l}\end{array}$

$\begin{array}{rcl}m& =& -l(\alpha +\beta )\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}$

$\begin{array}{rcl}\mathit{P}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{r}\mathit{o}\mathit{o}\mathit{t}\mathbf{}& \mathbf{=}& \mathbf{}\frac{\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{t}}{\mathbf{c}\mathbf{o}\mathbf{e}\mathbf{f}\mathbf{f}\mathbf{i}\mathbf{e}\mathbf{c}\mathbf{i}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}{\mathbf{x}}^{\mathbf{2}}}\\ \alpha \beta & =& \frac{n}{l}\\ n& =& l\alpha \beta \end{array}$

Step2. Substitute the values of $m,n$ in given equation.

$m^3+l^{2}n+{\ln }^2=3lmn$

Put the value of $m$ and $n$ in the above equation

${\{-l(\alpha +\beta \left)\right\}}^3+l^3\alpha \beta +l^3{\alpha }^2{\beta }^2=-3l^3(\alpha +\beta )\alpha \beta$

$\Rightarrow$ ${(\alpha +\beta )}^3-\alpha \beta -{\alpha }^2{\beta }^2=3(\alpha +\beta )\alpha \beta$

$\Rightarrow$${\alpha }^3+{\beta }^3+3(\alpha +\beta )\alpha \beta -\alpha \beta -{\alpha }^2{\beta }^2=3(\alpha +\beta )\alpha \beta$

$\Rightarrow$ ${\alpha }^3+{\beta }^3-\alpha \beta -{\alpha }^2{\beta }^2=0$

$\Rightarrow$ ${\alpha }^3-\alpha \beta +{\beta }^3-{\alpha }^2{\beta }^2=0$

$\Rightarrow$ $\alpha ({\alpha }^2-\beta )-{\beta }^2({\alpha }^2-\beta )=0$

$\Rightarrow$ $({\alpha }^2-\beta )(\alpha -{\beta }^2)=0$

$\Rightarrow {\alpha }^2=\beta or\alpha ={\beta }^2$

$\mathbf{\therefore }\alpha ={\beta }^2$

Hence, option $\left(A\right)$ is correct answer