Question

Find all values of $\alpha$ for which the system of inequality.
$x^2+2x+\alpha ⩽0,x^2-4x-6\alpha ⩽0$ has a unique solution.

Answer 1

$x^2+2x+\alpha \le 0,x^2-4x-6\alpha \le 0$

$(x-\frac{-2\pm \sqrt{4-4\alpha }}{2})\le 0$-Equation 1

$(x-\frac{4\pm \sqrt{16+24\alpha }}{2})\le 0$-Equation 2

For unique solution, $D\equiv 0,$

For Equation 1, $4-4\alpha =0$

$\Rightarrow \alpha =1$

For Equation 2, $16+24\alpha =0$

$\Rightarrow \alpha =\frac{-16}{24}$

$\Rightarrow \alpha =\frac{-2}{3}$