If the equation $x^2+ax+b=0$ has real roots $\alpha$ and $\beta$, and the equation $x^2+cx+d=0$ has real roots $\alpha +1$ and $\beta +1$, then the value of $(a-c{)}^2$ is equal to:

Solution

1. From the first equation, $\alpha +\beta =-a$.
2. From the second equation, $(\alpha +1)+(\beta +1)=-c$.
3. This implies $\alpha +\beta +2=-c$.
4. Substituting $-a$ for $\alpha +\beta$, we get $-a+2=-c$, so $c-a=2$.
5. Thus $$\begin{array}{rl}(a-c{)}^2& =(-2{)}^2\\ & =4\end{array}$$.
6. Hence the answer is (D).