Let $a,b,c$ be nonzero real numbers such that $a+b+c\neq 0$ and $4a-2b+c\neq 0$. If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$, then which of the following equations has the roots $\frac{\alpha+2}{\alpha-1}$ and $\frac{\beta+2}{\beta-1}$?

Let the new root be

$y=\frac{x+2}{x-1}$

Now,

$y(x-1)=x+2$

$xy-y=x+2$

$x(y-1)=y+2$

$x=\frac{y+2}{y-1}$

Since $x$ is a root of

$ax^2+bx+c=0$

Put $x=\frac{y+2}{y-1}$.

$a\left(\frac{y+2}{y-1}\right)^2+b\left(\frac{y+2}{y-1}\right)+c=0$

Multiplying by $(y-1)^2$,

$a(y+2)^2+b(y+2)(y-1)+c(y-1)^2=0$

Now expand:

$a(y^2+4y+4)+b(y^2+y-2)+c(y^2-2y+1)=0$

So,

$(a+b+c)y^2+(4a+b-2c)y+(4a-2b+c)=0$

Therefore, the required equation is

$(a+b+c)x^2+(4a+b-2c)x+(4a-2b+c)=0$