Find the value of α for which the equation

Given:

$(\alpha-12) x^{2}+2(\alpha-12) x+2=0$

Here,

$a=(\alpha-12), b=2(\alpha-12)$ and $c=2$

It is given that the roots of the equation are equal; therefore, we have:

$D=0$

$\Rightarrow\left(b^{2}-4 a c\right)=0$

$\Rightarrow\{2(\alpha-12)\}^{2}-4 \times(\alpha-12) \times 2=0$

$\Rightarrow 4\left(\alpha^{2}-24 \alpha+144\right)-8(\alpha-12)=0$

$\Rightarrow 4 \alpha^{2}-96 \alpha+576-8 \alpha+96=0$

$\Rightarrow 4 \alpha^{2}-104 \alpha+672=0$

$\Rightarrow \alpha^{2}-26 \alpha+168=0$

$\Rightarrow \alpha^{2}-14 \alpha-12 \alpha+168=0$

$\Rightarrow \alpha(\alpha-14)-12(\alpha-14)=0$

$\Rightarrow(\alpha-14)(\alpha-12)=0$

$\therefore \alpha=14$ or $\alpha=12$

If the value of $\alpha$ is 12, the given equation becomes non-quadratic.

Therefore, the valueof $\alpha$ will be 14 for the equation to have equal roots.