If the quadratic equation formed by eliminating $x$ from $x^2+\alpha x+\beta=0$ and $x y+l(x+y)+m=0$ has the…
If the quadratic equation formed by eliminating $x$ from $x^2+\alpha x+\beta=0$ and $x y+l(x+y)+m=0$ has the same roots as that of the given quadratic equation, then the set of values of $\beta$ is
$\{m, \alpha l-m\}$
$\{m, l+m\}$
$\{m, \alpha\rfloor+m\}$
$\{m, I-m\}$
Solution
The new quadratic equation formed by after elimination of ' $x$ ' is,
$
\begin{aligned}
& \left(\frac{l y+m}{y+l}\right)^2-\alpha\left(\frac{l y+m}{y+l}\right)+\beta=0 \\
& \Rightarrow\left(l^2 y^2+m^2+2 l m y\right)-\alpha(l y+m)(y+l) \\
& \quad+\beta(y+l)^2=0 \\
& \Rightarrow\left(l^2-\alpha l+\beta\right) y^2+\left(2 l m-\alpha l^2-\alpha m+2 \beta l\right) \\
& y+\left(m^2-\alpha m l+\beta l^2\right)=0
\end{aligned}
$
$\because$ Eq. (i) and $x^2+\alpha x+\beta=0$ has same roots, so
$
\begin{aligned}
\frac{l^2-\alpha l+\beta}{1} & =\frac{2 l m-\alpha l^2-\alpha m+2 \beta l}{\alpha} \\
& =\frac{m^2-\alpha m l+\beta l^2}{\beta}
\end{aligned}
$
So, $\quad \beta l^2-\beta \alpha l+\beta^2=m^2-\alpha m l+\beta l^2$
$
\begin{array}{rrrl}
\Rightarrow & \beta^2-\beta(l \alpha)-m(m-\alpha l) & =0 \\
\Rightarrow & \beta^2-\beta m+\beta(m-\alpha l)-m(m-\alpha l) & =0 \\
\Rightarrow & \beta(\beta-m)+(m-\alpha l)(\beta-m) & =0 \\
\Rightarrow & & \beta=m, \alpha l-m
\end{array}
$