If $\alpha$ and $\beta$ are two double roots of $x^2+3(a+3) x-9 a=0$ for different values of…
If $\alpha$ and $\beta$ are two double roots of $x^2+3(a+3) x-9 a=0$ for different values of $\alpha(\alpha\gt\beta)$. then the minimum value of $x^2+\alpha x-\beta=0$ is
$\frac{69}{4}$
$-\frac{69}{4}$
$-\frac{35}{4}$
$\frac{35}{4}$
Solution
For double roots $D=b^2-4 a c=0$
$\Rightarrow 9(a+3)^2+36 a=0 \Rightarrow a^2+10 a+9=0$
$\Rightarrow(a+9)(a+1)=0 \Rightarrow a=-1,-9$
For $a=-1$, equation becomes $x^2+6 x+9=0 \Rightarrow x=-3$
For $a=-9$ equation becomes
$x^2-18 x+81=0 \Rightarrow x=9 \quad \therefore \alpha=9, \beta=-3(\because \alpha\gt\beta)$
Now, given equation becomes
$x^2+9 x+3=0 \Rightarrow\left(x+\frac{9}{2}\right)^2-\frac{69}{4}=0$
We know that $\left(x+\frac{9}{2}\right)^2 \geq 0 \quad \therefore\left(x+\frac{9}{2}\right)^2-\frac{69}{4} \geq-\frac{69}{4}$
Hence, minimum value $=\frac{-69}{4}$